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PEARSON EDEXGEL INTERNATIONAL AS / A LEVEL

PHYSICS STUDENT BOOK 1 物理学生读物 1

MILES HUDSON 迈尔斯·哈德森

PDF COMPILED BY SAAD 由 SAAD 编译的 PDF

PEARSON EDEXCEL INTERNATIONAL AS/A LEVEL PHYSICS
培生爱德思国际 AS/A LEVEL 物理

Student Book 1 学生用书 1
Miles Hudson 迈尔斯·哈德森
Published by Pearson Education Limited, 80 Strand, London, WC2R ORL.
由 Pearson Education Limited, 80 Strand, London, WC2R ORL 出版。

www.pearsonglobalschools.com
Copies of official specifications for all Pearson Edexcel qualifications may be found on the website: https://qualifications.pearson.com
所有 Pearson Edexcel 资格证书的官方规格副本可在网站上找到:https://qualifications.pearson.com
Text © Pearson Education Limited 2018
文 © 培生教育有限公司 2018

Designed by Tech-Set Ltd, Gateshead, UK
由英国盖茨黑德的 Tech-Set Ltd 设计

Edited by Kate Blackham and Jane Read
编辑:Kate Blackham 和 Jane Read

Typeset by Tech-Set Ltd, Gateshead, UK
由英国盖茨黑德的 Tech-Set Ltd 排版

Original illustrations © Pearson Education Limited 2018
原创插图 © 培生教育有限公司 2018

Cover design by Pearson Education Limited 2018
封面设计:培生教育有限公司 2018

Picture research by Aptara, Inc
图片研究由 Aptara, Inc 提供

Cover photo © RUSSELL CROMAN/SCIENCE PHOTO LIBRARY
封面照片 © RUSSELL CROMAN/SCIENCE PHOTO LIBRARY

Inside front cover photo: Dmitry Lobanov
封面内页照片:Dmitry Lobanov

The right of Miles Hudson to be identified as author of this work has been asserted by him in accordance with the Copyright, Designs and Patents Act 1988.
Miles Hudson 有权根据 1988 年《版权、设计和专利法》主张被确定为本作品的作者。
First published 2018 首次发布于 2018 年
21201918
10987654321
British Library Cataloguing in Publication Data
大英图书馆出版物数据编目

A catalogue record for this book is available from the British Library
本书的目录记录可从大英图书馆获得

ISBN 9781292244877 国际标准书号9781292244877
All rights reserved. No part of this publication may be reproduced in any form or by any means (including photocopying or storing it in any medium by electronic means and whether or not transiently or incidentally to some other use of this publication) without the written permission of the copyright owner, except in accordance with the provisions of the Copyright, Designs and Patents Act 1988 or under the terms of a licence issued by the Copyright Licensing Agency, Barnard’s Inn, 86 Fetter Lane, London EC4A 1EN (www.cla.co.uk). Applications for the copyright owner’s written permission should be addressed to the publisher.
保留所有权利。未经版权所有者的书面许可,不得以任何形式或任何方式复制本出版物的任何部分(包括通过电子方式将其影印或存储在任何媒体中,无论是否暂时或偶然地用于本出版物的其他用途),除非符合版权的规定, 1988 年《外观设计和专利法》或根据版权许可机构颁发的许可条款,Barnard's Inn, 86 Fetter Lane, London EC4A 1EN (www.cla.co.uk)。版权所有者书面许可的申请应提交给出版商。
Printed by Neografia in Slovakia
由 Neografia 在斯洛伐克印刷

Endorsement statement 认可声明

In order to ensure that this resource offers high-quality support for the associated Pearson qualification, it has been through a review process by the awarding body. This process confirmed that this resource fully covers the teaching and learning content of the specification at which it is aimed. It also confirms that it demonstrates an appropriate balance between the development of subject skills, knowledge and understanding, in addition to preparation for assessment.
为了确保该资源为相关的 Pearson 资格认证提供高质量的支持,它已通过颁奖机构的审查过程。这个过程证实了该资源完全涵盖了其所针对的规范的教学内容。它还证实,除了为评估做准备外,它还展示了学科技能、知识和理解的发展之间的适当平衡。
Endorsement does not cover any guidance on assessment activities or processes (e.g. practice questions or advice on how to answer assessment questions) included in the resource, nor does it prescribe any particular approach to the teaching or delivery of a related course.
认可不包括资源中包含的任何评估活动或过程的指导(例如练习题或关于如何回答评估问题的建议),也不规定相关课程的教学或交付的任何特定方法。
While the publishers have made every attempt to ensure that advice on the qualification and its assessment is accurate, the official specification and associated assessment guidance materials are the only authoritative source of information and should always be referred to for definitive guidance.
虽然出版商已尽一切努力确保有关资格认证及其评估的建议准确无误,但官方规范和相关的评估指导材料是唯一的权威信息来源,应始终参考以获得明确的指导。
Pearson examiners have not contributed to any sections in this resource relevant to examination papers for which they have responsibility.
Pearson 考官未为本资源中与其负责的试卷相关的任何部分做出贡献。
Examiners will not use endorsed resources as a source of material for any assessment set by Pearson. Endorsement of a resource does not mean that the resource is required to achieve this Pearson qualification, nor does it mean that it is the only suitable material available to support the qualification, and any resource lists produced by the awarding body shall include this and other appropriate resources.
考官不会使用认可的资源作为 Pearson 设置的任何评估的材料来源。认可资源并不意味着需要该资源才能获得此 Pearson 资格,也不意味着它是支持该资格的唯一合适材料,颁奖机构制作的任何资源清单均应包括此资源和其他适当资源。

Acknowledgements 确认

Logos 标志

Logo on page 168 from LIFEPAK® 1000 DEFIBRILLATOR brochure, http://www. physio-control.com/uploadedFiles/Physio85/Contents/Emergency_Medical_Care/ Products/Brochures/LP1000_Brochure 20w 20Rechargable 20Battery_3303851_C. pdf, copyright © 2012 Physio-Control, Inc.
LIFEPAK® 1000 除颤监护仪手册第 168 页的徽标,http://wwwphysio-control.com/uploadedFiles/Physio85/Contents/Emergency_Medical_Care/产品/宣传册/LP1000_Brochure 20w 20可充电20Battery_3303851_C. pdf,版权所有 © 2012 Physio-Control, Inc.

Tables 

Table on page 84 from Mountaineering: The Freedom of the Hills, 6th Edition, published by Mountaineers Books, Seattle and Quiller Publishing, Shrewsbury 148, copyright © 1997. Illustration and table reprinted with permission of the publisher; Table on page 84 adapted from ‘The 2013/14 catalogue of DMM International’, www.dmmwales.com. Reproduced with permission from DMM International Ltd; Table on page 144 from ‘Electrical Power Annual’ report, December 2013, The United States Energy Information Association Source: U. S. Energy Information Administration (Dec 2013).
第 84 页的表格,摘自《登山:山丘的自由》,第 6 版,由西雅图登山者图书公司和 Quiller 出版社出版,什鲁斯伯里 148 号,版权所有 © 1997。插图和表格经出版商许可转载;第 84 页的表格改编自 www.dmmwales.com 的“DMM International 2013/14 目录”。经 DMM International Ltd 许可转载;第 144 页的表格摘自“电力年度”报告,2013 年 12 月,美国能源信息协会来源:美国能源信息署(2013 年 12 月)。

Text 发短信

Extract on page 54 from manufacture hockey goalkeeping equipment, 8 July 2014, http://www.obo.co.nz/the-o-lab, OBO. Reproduced by permission; Extract on page 70 from ‘What is a Plimsoll line?’, http://oceanservice.noaa.gov/facts/plimsoll-line. html. Source: NOAA’s National Ocean Service; Extract on page 96 from USGC FAQs, http://www.usgs.gov/faq. Source: U.S. Geological Survey Department of the Interior/USGS; Extract on page 114 adapted from ‘Tuning the Marimba Bar and Resonator’ http://lafavre.us/tuning-marimba, copyright © 2007 Jeffrey La Favre; Extract on page 126 from ‘Forensic Glass Comparison: Background Information Used in Data Interpretation’, publication number 09-04 of the Laboratory Division of the Federal Bureau of Investigation, Vol.11, No 2 (Maureen Bottrell),April 2009, Source: Federal Bureau of Investigation, Quantico, Virginia, USA; Extract on page 168 from LIFEPAK® 1000 DEFIBRILLATOR brochure, pp. 56-57, http://www. physiocontrol.com /uploadedFiles/Physio85/Contents/Emergency_Medical_ Care/Products/Brochures/LP1000_Brochure%20w%20Rechargable%20 Battery_3303851_C.pdf, copyright © 2012 Physio-Control, Inc. ; Extract on page 190 from ‘Rover Team Working to Diagnose Electrical Issue’, 2 November 2013, http://www.jpl.nasa.gov/news/news.php?feature=3958, Source: NASA.
第 54 页摘自制造曲棍球守门员设备,2014 年 7 月 8 日,http://www.obo.co.nz/the-o-lab,OBO。经许可转载;摘自第 70 页的“什么是 Plimsoll 线?”,http://oceanservice.noaa.gov/facts/plimsoll-line。html 的 URL 中。资料来源:NOAA 国家海洋局;第 96 页摘自 USGC 常见问题解答,http://www.usgs.gov/faq。资料来源:美国地质调查局内政部/USGS;第 114 页摘录自“调整马林巴琴杆和共鸣器”http://lafavre.us/tuning-marimba,版权所有 © 2007 Jeffrey La Favre;第 126 页摘自“法医玻璃比较:数据解释中使用的背景信息”,联邦调查局实验室部门出版物编号 09-04,第 11 卷,第 2 期 (Maureen Bottrell),2009 年 4 月,来源:美国弗吉尼亚州匡蒂科联邦调查局;摘自 LIFEPAK® 1000 除颤监护仪手册第 168 页,第 56-57 页,http://wwwphysiocontrol.com /uploadedFiles/Physio85/Contents/Emergency_Medical_ Care/Products/Brochures/LP1000_Brochure%20w%20Rechargable%20 Battery_3303851_C.pdf,版权所有 © 2012 Physio-Control, Inc. ;摘自“漫游者团队致力于诊断电气问题”,2013 年 11 月 2 日,http://www.jpl.nasa.gov/news/news.php?feature=3958,来源:NASA。
Every effort has been made to contact copyright holders of material reproduced in this book. Any omissions will be rectified in subsequent printings if notice is given to the publishers.
已尽一切努力联系本书中转载材料的版权所有者。如果通知出版商,任何遗漏将在后续印刷中得到纠正。
For Photo and Figure Acknowedgements please see page 214
有关照片和图表致谢,请参见第 214 页

COURSE STRUCTURE … iv 课程结构 ...四
ABOUT THIS BOOK … vi
关于这本书 ...六

PRACTICAL SKILLS … viii 实用技能 ...八
ASSESSMENT OVERVIEW … x x x\mathbf{x}
评估概述 ... x x x\mathbf{x}

WORKING AS A PHYSICIST … 2
作为一名物理学家......2

TOPIC 1 … 8 主题 1 ...8
TOPIC 2 … 58 主题 2 ...58
TOPIC 3 … 88 主题 3 ...88
TOPIC 4 … 148 主题 4 ...148
MATHS SKILLS … 194 数学技能 ...194
PREPARING FOR YOUR EXAMS … 200
准备考试 ...200

SAMPLE EXAM ANSWERS … 202
样本考试答案 ...202

COMMAND WORDS … 205 命令词 ...205
GLOSSARY … 207 词汇表。。。207
INDEX … 211 指数。。。211

WORKING AS A PHYSICIST 作为物理学家工作

1 STANDARD UNITS IN PHYSICS
1 物理学标准单位
2 ESTIMATION 2 估计

TOPIC 1 MECHANICS 主题 1:机制

1A MOTION 1A 运动
1 VELOCITY AND ACCELERATION
1 速度和加速度
2 MOTION GRAPHS 2 个运动图
3 ADDING FORCES 15 3 加力 15
4 MOMENTS 17 4 时刻 17
5 NEWTON’S LAWS 5 牛顿定律
OF MOTION 运动
6 KINEMATICS EQUATIONS 23
6 运动学方程 23

7 RESOLVING VECTORS 26 7 解析矢量 26
8 PROJECTILES 28 8 弹道 28
THINKING BIGGER: 放眼大处:
THE BATTLE OF AGRA 30
阿格拉战役 30

EXAM PRACTICE 32 考试练习 32
1B ENERGY 34 1B 能源 34
1 GRAVITATIONAL POTENTIAL AND KINETIC ENERGIES36
1 引力势和动能ENERGIES36
2 WORK AND POWER 39
2 工作和力量 39

THINKING BIGGER: 放眼大处:
THE MECHANICS 机制
OF SOCCER42
EXAM PRACTICE … 44 考试练习 ...44
1C MOMENTUM 1C 动量 46
1 MOMENTUM 1 动量 48
2 CONSERVATION OF 2 保护
LINEAR MOMENTUM 线性动量 50
1C MOMENTUM 46 1 MOMENTUM 48 2 CONSERVATION OF LINEAR MOMENTUM 50| 1C MOMENTUM | 46 | | :--- | :--- | | 1 MOMENTUM | 48 | | 2 CONSERVATION OF | | | LINEAR MOMENTUM | 50 |
THINKING BIGGER: 放眼大处:
SAVING HOCKEY GOALKEEPERS
拯救曲棍球守门员

EXAM PRACTICE … 56 考试练习 ...56
TOPIC 2 主题 2
MATERIALS 材料
2A FLUIDS … 58 2A 流体 ...58
1 FLUIDS, DENSITY AND UPTHRUST … 60
1 流体、密度和上推力 ...60

2 FLUID MOVEMENT … 63
2 流畅的运动 ...63

3 VISCOSITY … 65 3 粘度 ...65
4 TERMINAL VELOCITY … 67
4 终端速度 ...67

THINKING BIGGER: 放眼大处:
THE PLIMSOLL LINE … 70
普利姆索尔线 ...70

EXAM PRACTICE … 72 考试练习 ...72
2B SOLID MATERIAL PROPERTIES … 74
2B 固体材料特性 ...74

1 HOOKE’S LAW … 76
1 胡克定律 ...76

2 STRESS, STRAIN AND THE YOUNG MODULUS … 79
2 应力、应变和杨氏模量 ...79

3 STRESS-STRAIN GRAPHS … 81
3 应力-应变图 ...81

THINKING BIGGER 放眼更远
GET ROPED IN … 84
加入 ...84

EXAM PRACTICE … 86 考试练习 ...86

TOPIC 3
WAVES AND PARTICLE NATURE OF LIGHT
主题 3 光的波和粒子性质

3A BASIC WAVES 88 3A 基本波 88
1 WAVE BASICS 90 1 WAVE 基础知识 90
2 WAVE TYPES 93 2 波型 93
THINKING BIGGER: 放眼大处:
EARTHQUAKE 地震
96
EXAM PRACTICE … 98 考试练习 ...98
3B THE BEHAVIOUR OF WAVES … 100
3B 波浪的行为......100

1 WAVE PHASE AND SUPERPOSITION … 102
1 波相位和叠加 ...102

2 STATIONARY WAVES … 105
2 驻波 ...105

3 DIFFRACTION … 108 3 衍射 ...108
4 WAVE INTERFERENCE … 111
4 波干扰111

THINKING BIGGER THE MARIMBA … 114
想得更大 马林巴琴 ...114

EXAM PRACTICE … 116 考试练习 ...116
3C MORE WAVE PROPERTIES OF LIGHT … 118
3C 更多光的波浪特性118

1 REFRACTION … 120 1 折射 ...120
2 TOTAL INTERNAL REFLECTION … 122
2 全内反射 ...122

3 POLARISATION … 124 3 极化 ...124
THINKING BIGGER: GLASS FORENSICS … 126
放眼大处:玻璃法医 ...126

EXAM PRACTICE … 128 考试练习 ...128

3D QUANTUM PHYSICS 3D 量子物理学

1 WAVE-PARTICLE 1 波粒
DUALITY 二重性
2 THE PHOTOELECTRIC EFFECT
2 光电效应
3 ELECTRON DIFFRACTION AND INTERFERENCE
3 电子衍射和干涉
4 ATOMIC ELECTRON ENERGIES
4 原子电子能
THINKING BIGGER: 放眼大处:
SOLAR CELLS TO POWER 太阳能电池供电
THE USA? 美国?
EXAM PRACTICE 考试练习
TOPIC 4 ELECTRIC CIRCUITS
主题 4 电路

4A ELECTRICAL QUANTITIES 4A 电量

1 ELECTRIC CURRENT 1 电流
2 ELECTRICAL ENERGY TRANSFER
2 电能传输
3 CURRENT AND VOLTAGE RELATIONSHIPS156
3 电流和电压 RELATIONSHIPS156

4 RESISTIVITY … 159 4 电阻率 ...159
5 CONDUCTION AND RESISTANCE … 161
5 传导和电阻161

6 SEMICONDUCTORS … 164 6 半导体 ...164
THINKING BIGGER: 放眼大处:
SHOCKING STUFF … 168 令人震惊的东西......168
EXAM PRACTICE … 170 考试练习 ...170
4B COMPLETE 4B 完成
ELECTRICAL CIRCUITS … 172
电路。。。172

1 SERIES AND PARALLEL CIRCUITS … 174
1 串联和并联电路 ...174

2 ELECTRICAL CIRCUIT RULES … 178
2 电路规则178

3 POTENTIAL DIVIDERS … 180
3 个可能的分隔器 ...180

4 EMF AND INTERNAL RESISTANCE … 184
4 电动势和内阻 ...184

5 POWER IN ELECTRIC CIRCUITS … 186
5 电路中的电源 ...186

THINKING BIGGER: 放眼大处:
CURIOUS VOLTAGE DROP … 190
奇怪的电压降 ...190

EXAM PRACTICE … 192 考试练习 ...192
MATHS SKILLS … 194 数学技能 ...194
PREPARING FOR YOUR EXAMS … 200
准备考试 ...200

SAMPLE EXAM ANSWERS … 202
样本考试答案 ...202

COMMAND WORDS … 205 命令词 ...205
GLOSSARY … 207 词汇表。。。207
INDEX … 211 指数。。。211

ABOUT THIS BOOK 关于本书

This book is written for students following the Pearson Edexcel International Advanced Subsidiary (IAS) Physics specification. This book covers the full IAS course and the first year of the International A Level (IAL) course.
本书是为遵循 Pearson Edexcel International Advanced Subsidiary (IAS) 物理规范的学生编写的。本书涵盖了完整的 IAS 课程和国际 A Level (IAL) 课程的第一年。
The book contains full coverage of IAS units (or exam papers) 1 and 2. Each unit in the specification has two topic areas. The topics in this book, and their contents, fully match the specification. You can refer to the Assessment Overview on page X X XX for further information. Students can prepare for the written Practical Skills Paper (unit 3) by using the IAL Physics Lab Book (see page viii of this book).
本书全面涵盖 IAS 单元(或试卷)1 和 2。规范中的每个单元都有两个主题区域。本书中的主题及其内容完全符合规范。有关更多信息,请参阅页面上 X X XX 的评估概述。学生可以使用 IAL 物理实验书(见本书第 viii 页)为书面实践技能试卷(第 3 单元)做准备。
Each Topic is divided into chapters and sections to break the content down into manageable chunks. Each section features a mix of learning and activities.
每个主题都分为章节和部分,以将内容分解为可管理的块。每个部分都包含学习和活动。
Learning objectives 学习目标
Each chapter starts with a list of key assessment objectives.
每章都以关键评估目标列表开始。

1 A 1 VELOCITY AND ACGELERATION
1 A 1 速度和加速度

Mowemenn is fundamental to the denctioning of our univesse. Whetber you are rurring to catch a
Mowemenn 是我们大学的基础。你急于赶上

RATE OF MOVEMENT 移动速率
One of the simplest thans we can measure is how tas an abject is moving You can cakculate an arestain distance
我们可以测量的最简单的方法之一是 tas an abject 是如何移动的 You can cakculate a arestain distance

Howews the calculstion for powed will only tell you how fast an nbigect is moring Often it is also
Howews powed 的计算只会告诉你 nbigect 的 moring 有多快 通常也是



OR v = s t v = Δ s M v = s t v = Δ s M quad^(v=(s)/(t))v=(Delta s)/(M)\quad{ }^{v=\frac{s}{t}} \mathrm{v}=\frac{\Delta s}{M}  v = s t v = Δ s M v = s t v = Δ s M quad^(v=(s)/(t))v=(Delta s)/(M)\quad{ }^{v=\frac{s}{t}} \mathrm{v}=\frac{\Delta s}{M}

the messurement is reseredet to os a a colalar quasunity The
测量是 reseredet to os a a colalar quasunity

scalar. aistarce is a scalar and depplacement is a vectoc.
标量。aistarce 是标量,而 depplacement 是 vectoc。

Specification reference The exact specification references covered in the section are provided.
规范参考 提供了本节中涵盖的确切规范参考。
Subject vocabulary 学科词汇
Key terms are highlighted in blue in the text. Clear definitions are provided at the end of each section for easy reference, and are also collated in a glossary at the back of the book.
关键术语在文本中以蓝色突出显示。每个部分的末尾都提供了清晰的定义,以便于参考,并且还整理在本书后面的词汇表中。

Exam hints 考试提示

Tips on how to answer examstyle questions and guidance for exam preparation. Worked examples also show you how to work through questions and set out calculations.
有关如何回答 Examstyle 问题的提示和考试准备指南。工作示例还向您展示了如何解决问题和设置计算。

Checkpoint 检查站

Questions at the end of each section check understanding of the key learning points in each chapter.
每部分末尾的问题检查对每章中关键学习点的理解。
These help you focus your learning and avoid common errors.
这些可以帮助您集中精力学习并避免常见错误。
Did you know? 您知道吗?
Interesting facts help you to remember the key concepts.
有趣的事实可以帮助您记住关键概念。
Your learning, chapter by chapter, is always put in context:
您的学习,逐章,始终放在上下文中:
  • Links to other areas of Physics include previous knowledge that is built on in the topic, and future learning that you will cover later in your course.
    指向物理学其他领域的链接包括该主题所建立的先前知识,以及您将在课程后面介绍的未来学习。
  • A checklist details maths knowledge required. If you need to practise these skills, you can use the Maths Skills reference at the back of the book as a starting point.
    清单详细说明了所需的数学知识。如果您需要练习这些技能,您可以使用书后面的 Maths Skills 参考资料作为起点。

TOPIC 1 MECHANICS 主题 1:机制

1A MOTION 1A 运动








MathS sKILL FOR THIS CHAPTER
MathS sKILL 本章

Unit of messuremem ieg are nemoan n n nn






1A THINKING BIGGER 1A 思考更大
THE BATTLE OF AGRA 阿格拉战役


Thinking Bigger 放眼大处

At the end of each topic, there is an opportunity to read and work with real-life research and writing about science. The activities help you to read real-life material that’s relevant to your course, analyse how scientists write, think critically and consider how different aspects of your learning piece together.
在每个主题的末尾,都有机会阅读和处理现实生活中的研究以及有关科学的写作。这些活动可帮助您阅读与您的课程相关的现实生活中的材料,分析科学家的写作方式,批判性地思考,并考虑学习的不同方面如何组合在一起。

Skills 技能

These sections will help you develop transferable skills, which are highly valued in further study and the workplace.
这些部分将帮助您培养可转移的技能,这些技能在进一步学习和工作场所中受到高度重视。

Exam Practice 考试练习

Exam-style questions at the end of each chapter are tailored to the Pearson Edexcel specification to allow for practice and development of exam writing technique. They also allow for practice responding to the command words used in the exams (see the command words glossary at the back of this book).
每章末尾的考试式问题是根据 Pearson Edexcel 规范量身定制的,以便练习和发展考试写作技巧。它们还允许练习回答考试中使用的命令词(请参阅本书后面的命令词词汇表)。

1A EXAM PRACTICE 1A 考试练习


PRACTICAL SKILLS 实践技能

Practical work is central to the study of physics. The Pearson Edexcel International Advanced Subsidiary (IAS) Physics specification includes eight Core Practicals that link theoretical knowledge and understanding to practical scenarios.
实践工作是物理学研究的核心。Pearson Edexcel International Advanced Subsidiary (IAS) 物理规范包括八项核心实践,将理论知识和理解与实际场景联系起来。
Your knowledge and understanding of practical skills and activities will be assessed in all examination papers for the IAS Level Physics qualification.
您对实践技能和活动的知识和理解将在 IAS Level Physics 资格的所有试卷中进行评估。
  • Papers 1 and 2 will include questions based on practical activities, including novel scenarios.
    试卷 1 和 2 将包括基于实践活动的问题,包括新场景。
  • Paper 3 will test your ability to plan practical work, including risk management and selection of apparatus.
    试卷 3 将测试您规划实际工作的能力,包括风险管理和设备选择。
In order to develop practical skills, you should carry out a range of practical experiments related to the topics covered in your course. Further suggestions in addition to the Core Practicals are included below.
为了培养实践技能,您应该进行一系列与课程涵盖的主题相关的实践实验。除了核心实践之外,下面还包括更多建议。
STUDENT BOOK TOPIC 学生用书主题 IAS CORE PRACTICALS IAS 核心实践
TOPIC 1 MECHANICS 主题 1:机制 CP1 Determine the acceleration of a freelyfalling object
确定自由落体物体的加速度
 主题 2 材料
TOPIC 2
MATERIALS
TOPIC 2 MATERIALS| TOPIC 2 | | :--- | | MATERIALS |
CP2 Use a falling-ball method to determine the viscosity of a liquid
使用落球法测定液体的粘度
CP3 Determine the Young modulus of a material
确定材料的杨氏模量

主题 3:波和光的粒子性质
TOPIC 3
WAVES AND THE PARTICLE NATURE OF LIGHT
TOPIC 3 WAVES AND THE PARTICLE NATURE OF LIGHT| TOPIC 3 | | :--- | | WAVES AND THE PARTICLE NATURE OF LIGHT |
CP4 Determine the speed of sound in air using a two-beam oscilloscope, signal generator, speaker and microphone
使用双波束示波器、信号发生器、扬声器和麦克风确定空气中的声速
CP5 Investigate the effects of length, tension and mass per unit length on the frequency of a vibrating string or wire
研究长度、张力和每单位长度的质量对振弦或线材频率的影响
CP6 Determine the wavelength of light from a laser or other light source using a diffraction grating
使用衍射光栅确定来自激光或其他光源的光的波长

主题 4 电路
TOPIC 4
ELECTRIC CIRCUITS
TOPIC 4 ELECTRIC CIRCUITS| TOPIC 4 | | :--- | | ELECTRIC CIRCUITS |
CP7 Determine the electrical resistivity of a material
确定材料的电阻率
CP8 Determine the e.m.f. and internal resistance of an electrical cell
确定电池的电动势和内阻
STUDENT BOOK TOPIC IAS CORE PRACTICALS TOPIC 1 MECHANICS CP1 Determine the acceleration of a freelyfalling object "TOPIC 2 MATERIALS" CP2 Use a falling-ball method to determine the viscosity of a liquid CP3 Determine the Young modulus of a material "TOPIC 3 WAVES AND THE PARTICLE NATURE OF LIGHT" CP4 Determine the speed of sound in air using a two-beam oscilloscope, signal generator, speaker and microphone CP5 Investigate the effects of length, tension and mass per unit length on the frequency of a vibrating string or wire CP6 Determine the wavelength of light from a laser or other light source using a diffraction grating "TOPIC 4 ELECTRIC CIRCUITS" CP7 Determine the electrical resistivity of a material CP8 Determine the e.m.f. and internal resistance of an electrical cell| STUDENT BOOK TOPIC | IAS CORE PRACTICALS | | | :---: | :---: | :---: | | TOPIC 1 MECHANICS | CP1 | Determine the acceleration of a freelyfalling object | | TOPIC 2 <br> MATERIALS | CP2 | Use a falling-ball method to determine the viscosity of a liquid | | | CP3 | Determine the Young modulus of a material | | TOPIC 3 <br> WAVES AND THE PARTICLE NATURE OF LIGHT | CP4 | Determine the speed of sound in air using a two-beam oscilloscope, signal generator, speaker and microphone | | | CP5 | Investigate the effects of length, tension and mass per unit length on the frequency of a vibrating string or wire | | | CP6 | Determine the wavelength of light from a laser or other light source using a diffraction grating | | TOPIC 4 <br> ELECTRIC CIRCUITS | CP7 | Determine the electrical resistivity of a material | | | CP8 | Determine the e.m.f. and internal resistance of an electrical cell |

UNIT 1 (TOPICS 1 AND 2) MECHANICS AND MATERIALS
第 1 单元(主题 1 和 2)力学和材料

Possible further practicals include:
可能的进一步实践包括:
  • Strobe photography or the use of a video camera to analyse projectile motion
    频闪摄影或使用摄像机分析弹丸运动
  • Determine the centre of gravity of an irregular rod
    确定不规则杆的重心
  • Investigate the conservation of momentum using light gates and air track
    使用光门和空气轨迹研究动量守恒
  • Hooke’s law and the Young modulus experiments for a variety of materials
    各种材料的胡克定律和杨氏模量实验

UNIT 2 (TOPICS 3 AND 4)
第 2 单元(主题 3 和 4)

WAVES AND ELECTRICITY 波浪和电
Possible further practicals include:
可能的进一步实践包括:
  • Estimating power output of an electric motor
    估算电动机的功率输出
  • Using a digital voltmeter to investigate the output of a potential divider and investigating current/ voltage graphs for a filament bulb, thermistor and diode
    使用数字电压表研究分压器的输出,并研究白炽灯泡、热敏电阻和二极管的电流/电压图
  • Determining the refractive index of solids and liquids, demonstrating progressive and stationary waves on a slinky
    测定固体和液体的折射率,在 slinky 上展示行波和驻波

In the Student Book, the Core Practical specification and Lab Book references are supplied in the relevant sections.
在 Student Book 中,相关部分提供了 Core Practical 规范和 Lab Book 参考资料。

Practical Skills 实践技能

Practical skills boxes explain techniques used in the Core Practicals, and also detail useful skills and knowledge gained in other related investigations.
实用技能框解释了核心实践中使用的技术,还详细介绍了在其他相关调查中获得的有用技能和知识。
This Student Book is accompanied by a Lab Book, which includes instructions and writing frames for the Core Practicals for students to record their results and reflect on their work.
这本学生用书附有一本实验本,其中包括核心实践的说明和写作框架,供学生记录他们的结果并反思他们的工作。

Practical skills checklists, practice questions and answers are also provided.
还提供实用技能清单、练习题和答案。

The Lab Book records can be used as preparation and revision for the Practical Skills Paper.
实验手册记录可用作实用技能试卷的准备和修订。

ASSESSMENT OVERVIEW 评估概述

The following tables give an overview of the assessment for Pearson Edexcel International Advanced Subsidiary course in Physics. You should study this information closely to help ensure that you are fully prepared for this course and know exactly what to expect in each part of the examination. More information about this qualification, and about the question types in the different papers, can be found on page 200 of this book.
下表概述了 Pearson Edexcel 国际物理高级辅助课程的评估。您应该仔细研究这些信息,以帮助确保您为本课程做好充分准备,并确切了解考试的每个部分会发生什么。有关此资格以及不同论文中问题类型的更多信息,请参见本书的第 200 页。
PAPER / UNIT 1 纸张 / 单元 1 PERCENTAGE OF IAS IAS 的百分比 PERGENTAGE OF IAL IAL 的持久性 MARK TIME AVAILABILITY

机械和材料 笔试卷 试卷代码 WPH11/01 外部设置和评分 由 Pearson Edexcel 进行单级参赛
MECHANICS AND MATERIALS
Written exam paper
Paper code
WPH11/01
Externally set and marked by Pearson Edexcel
Single tier of entry
MECHANICS AND MATERIALS Written exam paper Paper code WPH11/01 Externally set and marked by Pearson Edexcel Single tier of entry| MECHANICS AND MATERIALS | | :--- | | Written exam paper | | Paper code | | WPH11/01 | | Externally set and marked by Pearson Edexcel | | Single tier of entry |
40 % 40 % 40%40 \% 20 % 20 % 20%20 \% 80

1 小时 30 分钟
1 hour
30 minutes
1 hour 30 minutes| 1 hour | | :--- | | 30 minutes |

1 月、6 月和 10 月 第一次评估 : 2019 年 1 月
January, June and October
First assessment : January 2019
January, June and October First assessment : January 2019| January, June and October | | :--- | | First assessment : January 2019 |
PAPER / UNIT 2 纸张 / 单元 2 PERCENTAGE PERCENTAGE OF IAL IAL 的百分比 MARK TIME AVAILABILITY

WAVES AND ELECTRICITY 笔试卷 试卷代码 WPH12/01 外部设置和评分 由 Pearson Edexcel 单级参赛
WAVES AND ELECTRICITY
Written exam paper
Paper code
WPH12/01
Externally set and marked by Pearson Edexcel
Single tier of entry
WAVES AND ELECTRICITY Written exam paper Paper code WPH12/01 Externally set and marked by Pearson Edexcel Single tier of entry| WAVES AND ELECTRICITY | | :--- | | Written exam paper | | Paper code | | WPH12/01 | | Externally set and marked by Pearson Edexcel | | Single tier of entry |
40 % 40 % 40%40 \% 20 % 20 % 20%20 \% 80

1 小时 30 分钟
1 hour
30 minutes
1 hour 30 minutes| 1 hour | | :--- | | 30 minutes |

2019 年 1 月、6 月和 10 月第一次评估
January, June and October
First assessment June 2019
January, June and October First assessment June 2019| January, June and October | | :--- | | First assessment June 2019 |
PAPER / UNIT 3 纸张 / 单元 3 PERCENTAGE
 IAL 的百分比
PERCENTAGE
OF IAL
PERCENTAGE OF IAL| PERCENTAGE | | :--- | | OF IAL |
MARK TIME AVAILABILITY

物理实用技能 1 笔试 试卷编号 WPH13/01 外部设置和评分 由 Pearson Edexcel 单级参赛
PRACTICAL SKILLS IN PHYSICS 1
Written examination
Paper code
WPH13/01
Externally set and marked by Pearson Edexcel
Single tier of entry
PRACTICAL SKILLS IN PHYSICS 1 Written examination Paper code WPH13/01 Externally set and marked by Pearson Edexcel Single tier of entry| PRACTICAL SKILLS IN PHYSICS 1 | | :--- | | Written examination | | Paper code | | WPH13/01 | | Externally set and marked by Pearson Edexcel | | Single tier of entry |
20 % 20 % 20%20 \% 10 % 10 % 10%10 \% 50 1 hour 20 minutes 1 小时 20 分钟

1 月、6 月和 10 月 第一次评估 : 2019 年 6 月
January, June and October
First assessment : June 2019
January, June and October First assessment : June 2019| January, June and October | | :--- | | First assessment : June 2019 |
PAPER / UNIT 1 PERCENTAGE OF IAS PERGENTAGE OF IAL MARK TIME AVAILABILITY "MECHANICS AND MATERIALS Written exam paper Paper code WPH11/01 Externally set and marked by Pearson Edexcel Single tier of entry" 40% 20% 80 "1 hour 30 minutes" "January, June and October First assessment : January 2019" PAPER / UNIT 2 PERCENTAGE PERCENTAGE OF IAL MARK TIME AVAILABILITY "WAVES AND ELECTRICITY Written exam paper Paper code WPH12/01 Externally set and marked by Pearson Edexcel Single tier of entry" 40% 20% 80 "1 hour 30 minutes" "January, June and October First assessment June 2019" PAPER / UNIT 3 PERCENTAGE "PERCENTAGE OF IAL" MARK TIME AVAILABILITY "PRACTICAL SKILLS IN PHYSICS 1 Written examination Paper code WPH13/01 Externally set and marked by Pearson Edexcel Single tier of entry" 20% 10% 50 1 hour 20 minutes "January, June and October First assessment : June 2019"| PAPER / UNIT 1 | PERCENTAGE OF IAS | PERGENTAGE OF IAL | MARK | TIME | AVAILABILITY | | :---: | :---: | :---: | :---: | :---: | :---: | | MECHANICS AND MATERIALS <br> Written exam paper <br> Paper code <br> WPH11/01 <br> Externally set and marked by Pearson Edexcel <br> Single tier of entry | $40 \%$ | $20 \%$ | 80 | 1 hour <br> 30 minutes | January, June and October <br> First assessment : January 2019 | | PAPER / UNIT 2 | PERCENTAGE | PERCENTAGE OF IAL | MARK | TIME | AVAILABILITY | | WAVES AND ELECTRICITY <br> Written exam paper <br> Paper code <br> WPH12/01 <br> Externally set and marked by Pearson Edexcel <br> Single tier of entry | $40 \%$ | $20 \%$ | 80 | 1 hour <br> 30 minutes | January, June and October <br> First assessment June 2019 | | PAPER / UNIT 3 | PERCENTAGE | PERCENTAGE <br> OF IAL | MARK | TIME | AVAILABILITY | | PRACTICAL SKILLS IN PHYSICS 1 <br> Written examination <br> Paper code <br> WPH13/01 <br> Externally set and marked by Pearson Edexcel <br> Single tier of entry | $20 \%$ | $10 \%$ | 50 | 1 hour 20 minutes | January, June and October <br> First assessment : June 2019 |

ASSESSMENT OBJECTIVES AND WEIGHTINGS
评估目标和权重

ASSESSMENT OBJECTIVE 评估目标 DESCRIPTION % IN IAS IAS 中的百分比 % IN IA2 IA2 中的百分比 % IN IAL 在 IAL 中 %
A01 Demonstrate knowledge and understanding of science
展示对科学的知识和理解
34 36 34 36 34-3634-36 29 31 29 31 29-3129-31 32 34 32 34 32-3432-34
A02

(a) 在熟悉和不熟悉的环境中应用知识和对科学的理解。 (b) 分析和评估科学信息以做出判断并得出结论。
(a) Application of knowledge and understanding of science in
familiar and unfamiliar contexts.
(b) Analysis and evaluation of scientific information to make
judgments and reach conclusions.
(a) Application of knowledge and understanding of science in familiar and unfamiliar contexts. (b) Analysis and evaluation of scientific information to make judgments and reach conclusions.| (a) Application of knowledge and understanding of science in | | :--- | | familiar and unfamiliar contexts. | | (b) Analysis and evaluation of scientific information to make | | judgments and reach conclusions. |
34 36 34 36 34-3634-36 33 36 33 36 33-3633-36 34 36 34 36 34-3634-36
A03

科学实验技能,包括数据和方法的分析和评估
Experimental skills in science, including analysis and evaluation
of data and methods
Experimental skills in science, including analysis and evaluation of data and methods| Experimental skills in science, including analysis and evaluation | | :--- | | of data and methods |
20 14 16 14 16 14-1614-16 11 14 11 14 11-1411-14
ASSESSMENT OBJECTIVE DESCRIPTION % IN IAS % IN IA2 % IN IAL A01 Demonstrate knowledge and understanding of science 34-36 29-31 32-34 A02 "(a) Application of knowledge and understanding of science in familiar and unfamiliar contexts. (b) Analysis and evaluation of scientific information to make judgments and reach conclusions." 34-36 33-36 34-36 A03 "Experimental skills in science, including analysis and evaluation of data and methods" 20 14-16 11-14| ASSESSMENT OBJECTIVE | DESCRIPTION | % IN IAS | % IN IA2 | % IN IAL | | :---: | :--- | :---: | :---: | :---: | | A01 | Demonstrate knowledge and understanding of science | $34-36$ | $29-31$ | $32-34$ | | A02 | (a) Application of knowledge and understanding of science in <br> familiar and unfamiliar contexts. <br> (b) Analysis and evaluation of scientific information to make <br> judgments and reach conclusions. | $34-36$ | $33-36$ | $34-36$ | | A03 | Experimental skills in science, including analysis and evaluation <br> of data and methods | 20 | $14-16$ | $11-14$ |

RELATIONSHIP OF ASSESSMENT OBJECTIVES TO UNITS
评估目标与单位的关系

UNIT NUMBER 单元号 ASSESSMENT OBJECTIVE 评估目标
A01 A02 (a) A02 (一) A02 (b) A02 (二) A03
UNIT 1 单元 1 17 18 17 18 17-1817-18 17 18 17 18 17-1817-18 4.5 5.5 4.5 5.5 4.5-5.54.5-5.5 0.0
UNIT 2 第 2 单元 17 18 17 18 17-1817-18 17 18 17 18 17-1817-18 4.5 5.5 4.5 5.5 4.5-5.54.5-5.5 0.0
UNIT 3 第 3 单元 0.0 0.0 0.0 20
TOTAL FOR INTERNATIONAL ADVANCED SUBSIDIARY
国际先进子公司合计
34 36 34 36 34-3634-36 34 36 34 36 34-3634-36 9 11 9 11 9-119-11 20
UNIT NUMBER ASSESSMENT OBJECTIVE A01 A02 (a) A02 (b) A03 UNIT 1 17-18 17-18 4.5-5.5 0.0 UNIT 2 17-18 17-18 4.5-5.5 0.0 UNIT 3 0.0 0.0 0.0 20 TOTAL FOR INTERNATIONAL ADVANCED SUBSIDIARY 34-36 34-36 9-11 20| UNIT NUMBER | ASSESSMENT OBJECTIVE | | | | | :---: | :---: | :---: | :---: | :---: | | | A01 | A02 (a) | A02 (b) | A03 | | UNIT 1 | $17-18$ | $17-18$ | $4.5-5.5$ | 0.0 | | UNIT 2 | $17-18$ | $17-18$ | $4.5-5.5$ | 0.0 | | UNIT 3 | 0.0 | 0.0 | 0.0 | 20 | | TOTAL FOR INTERNATIONAL ADVANCED SUBSIDIARY | $34-36$ | $34-36$ | $9-11$ | 20 |

WORKING AS A PHYSICIST 作为物理学家工作

Throughout your study of physics, you will develop knowledge and understanding of what it means to work scientifically. You will develop confidence in key scientific skills, such as handling and controlling quantities and units and making estimates. You will also learn about the ways in which the scientific community functions and how society as a whole uses scientific ideas.
在整个物理学习过程中,您将发展对科学工作意味着什么的知识和理解。您将培养对关键科学技能的信心,例如处理和控制数量和单位以及进行估算。您还将了解科学界的运作方式以及整个社会如何使用科学思想。

At the end of each chapter in this book, there is a section called Thinking Bigger. These sections are based broadly on the content of the chapter just completed, but they will also draw on your previous learning from earlier in the course or from your previous studies and point towards future learning and less familiar contexts. The Thinking Bigger sections will also help you to develop transferable skills. By working through these sections, you will:
在本书每一章的末尾,都有一个名为 Thinking bigger 的部分。这些部分大致基于刚刚完成的章节的内容,但它们也将借鉴您之前在课程早期或以前的学习中的学习,并指出未来的学习和不太熟悉的背景。Thinking Bigger 部分也将帮助您培养可转移的技能。通过完成这些部分,您将:
  • read real-life scientific writing in a variety of contexts and aimed at different audiences
    在各种环境中阅读针对不同受众的真实科学著作
  • develop an understanding of how the professional scientific community functions
    了解专业科学界的运作方式
  • learn to think critically about the nature of what you have read
    学会批判性地思考你所读内容的本质
  • understand the issues, problems and challenges that may be raised
    了解可能提出的问题、问题和挑战
  • gain practice in communicating information and ideas in an appropriate scientific way
    获得以适当的科学方式传达信息和想法的练习
  • apply your knowledge and understanding to unfamiliar contexts.
    将您的知识和理解应用于不熟悉的环境。
You will also gain scientific skills through the hands-on practical work that forms an essential part of your course. As well as understanding the experimental methods of the practicals, it is important that you develop the skills necessary to plan experiments and analyse and evaluate data. Not only are these very important scientific skills, but they will be assessed in your examinations.
您还将通过实践工作获得科学技能,这是您课程的重要组成部分。除了了解实践的实验方法外,您还必须培养规划实验以及分析和评估数据所需的技能。这些不仅是非常重要的科学技能,而且它们将在您的考试中进行评估。

MATHS SKILLS FOR PHYSICISTS
物理学家的数学技能

  • Recognise and make use of appropriate units in calculations (e.g. knowing the difference between base and derived units)
    在计算中识别并使用适当的单位(例如,了解基本单位和派生单位之间的区别)
  • Estimate results (e.g. estimating the speed of waves on the sea)
    估计结果(例如,估计海浪的速度)
  • Make order of magnitude calculations (e.g. estimating approximately what an answer should be before you start calculating, including using standard form)
    进行数量级计算(例如,在开始计算之前估计大致应该是什么答案,包括使用标准格式)
  • Use algebra to rearrange and solve equations (e.g. finding the landing point of a projectile)
    使用代数重新排列和求解方程(例如,找到弹丸的着陆点)
  • Recognise the importance of the straight line graph as an analysis tool for the verification and development of physical laws by experimentation (e.g. choosing appropriate variables to plot to generate a straight line graph with experimental data)
    认识到直线图作为分析工具的重要性,通过实验验证和发展物理定律(例如,选择合适的变量进行绘制,以生成包含实验数据的直线图)
  • Determine the slope and intercept of a linear graph (e.g. finding acceleration from a velocity-time graph)
    确定线性图的斜率和截距(例如,从速度-时间图中求加速度)
  • Calculate the area under the line on a graph (e.g. finding the energy stored in a stretched wire)
    计算图表上线下的面积(例如,找到存储在拉伸导线中的能量)
  • Use geometry and trigonometry (e.g. finding components of vectors)
    使用几何和三角函数(例如,查找向量的分量)

What prior knowledge do I need?
我需要什么先验知识?

  • Understanding and knowledge of physical facts, terminology, concepts, principles and practical techniques
    理解和了解物理事实、术语、概念、原理和实用技术
  • Applying the concepts and principles of physics, including the applications of physics, to different contexts
    将物理学的概念和原理(包括物理学的应用)应用于不同的环境
  • Appreciating the practical nature of physics and developing experimental and investigative skills based on the use of correct and safe laboratory techniques
    欣赏物理学的实用性,并在使用正确和安全的实验室技术的基础上发展实验和调查技能
  • Communicating scientific methods, conclusions and arguments using technical and mathematical language
    使用技术和数学语言交流科学方法、结论和论点
  • Consideration of the implications, including benefits and risks, of scientific and technological developments
    考虑科学和技术发展的影响,包括利益和风险
  • Understanding how society uses scientific knowledge to make decisions about the implementation of technological developments
    了解社会如何利用科学知识来做出有关技术发展实施的决策

    Understanding of how scientific ideas change over time, and the systems in place to validate these changes
    了解科学思想如何随时间变化,以及验证这些变化的系统

What will I study later?
我以后会学习什么?

  • Knowledge and understanding of further physical facts and terminology, deeper concepts, principles and more complex practical techniques
    了解和理解进一步的物理事实和术语、更深的概念、原理和更复杂的实践技术
  • Practical skills and techniques for some key physics experiments
    一些关键物理实验的实用技能和技巧
  • How to communicate information and ideas in appropriate ways using appropriate terminology
    如何使用适当的术语以适当的方式传达信息和想法
  • The implications of science and their associated benefits and risks
    科学的影响及其相关的益处和风险
  • The role of the scientific community in validating new knowledge and ensuring integrity
    科学界在验证新知识和确保完整性方面的作用
  • The ways in which society uses science to inform decision making
    社会利用科学为决策提供信息的方式

1 STANDARD UNITS IN PHYSICS
1 物理学标准单位

LEARNING OBJECTIVES 学习目标

  • Understand the distinction between base and derived quantities.
    了解基数和派生量之间的区别。

    ■ Understand the idea of a fixed system of units, and explain the SI system.
    ■ 理解固定单位制的概念,并解释 SI 制。

BASE AND DERIVED QUANTITIES
基础数量和派生数量

A fig A The international standard kilogram, officially known as the International Prototype Kilogram, is made from a mixture of platinum and iridium and is held at the Bureau International des Poids et Mesures in Paris. All other masses are defined by comparing with this metal cylinder.
A 图 A 国际标准千克,正式名称为国际原型千克,由铂和铱的混合物制成,保存在巴黎的国际计量局。所有其他质量都是通过与这个金属圆柱体进行比较来定义的。

Some measurements we make are of fundamental qualities of things in the universe. For example, the length of a pencil is a fundamental property of the object. Compare this with the pencil’s speed if you drop it. To give a value to the speed, we have to consider a distance moved, and the rate of motion over that distance - we also need to measure time and then do a calculation. You can see that there is a fundamental difference between the types of quantity that are length and speed. We call the length a base unit, whilst the speed is a derived unit. At present, the international scientific community uses seven base units, and from these all other units are derived. Some derived units have their own names. For example, the derived unit of force should be kg m s 2 kg m s 2 kgms^(-2)\mathrm{kg} \mathrm{m} \mathrm{s}^{-2}, but this has been named the newton (N). Other derived units do not get their own name, and we just list the base units that went together in deriving the quantity. For example, speed is measured in m s 1 m s 1 ms^(-1)\mathrm{m} \mathrm{s}^{-1}.
我们所做的一些测量是关于宇宙中事物的基本品质的。例如,铅笔的长度是对象的基本属性。将此值与铅笔掉落时的速度进行比较。为了给速度一个值,我们必须考虑移动的距离,以及该距离上的运动速率 - 我们还需要测量时间,然后进行计算。您可以看到 length 和 speed 的数量类型之间存在根本区别。我们将长度称为基本单位,而速度是派生单位。目前,国际科学界使用七个基本单位,所有其他单位都是从这些单位派生出来的。一些派生单位有自己的名称。例如,力的派生单位应该是 kg m s 2 kg m s 2 kgms^(-2)\mathrm{kg} \mathrm{m} \mathrm{s}^{-2} ,但这已被命名为牛顿 (N)。其他派生单位没有自己的名称,我们只列出在派生数量时一起出现的基本单位。例如,速度以 m s 1 m s 1 ms^(-1)\mathrm{m} \mathrm{s}^{-1} 为单位。
BASIC QUANTITY 基本数量 UNIT NAME 单元名称 UNIT SYMBOL 单位符号
mass 质量 kilogram 公斤 kg 公斤
time 时间 second 第二 s
length 长度 metre  m
electric current 电流 ampere 安培 A
temperature 温度 kelvin 开尔文 K
amount of substance 物质量 mole 鼹鼠 mol 摩尔
light intensity 光强度 candela 坎德拉 cd 光盘
BASIC QUANTITY UNIT NAME UNIT SYMBOL mass kilogram kg time second s length metre m electric current ampere A temperature kelvin K amount of substance mole mol light intensity candela cd| BASIC QUANTITY | UNIT NAME | UNIT SYMBOL | | :--- | :--- | :---: | | mass | kilogram | kg | | time | second | s | | length | metre | m | | electric current | ampere | A | | temperature | kelvin | K | | amount of substance | mole | mol | | light intensity | candela | cd |
table A A A\mathbf{A} The base units.
A A A\mathbf{A} 基本单位。

The choice of which quantities are the base ones is somewhat a matter of choice. The scientists who meet to decide on the standard unit system have chosen these seven. You might think that electric current is not a fundamental property, as it is the rate of movement of charge. So it could be derived from measuring charge and time. However, scientists had to pick what was fundamental and they chose current. This means that electric charge is a derived quantity found by multiplying current passing for a given time.
选择哪些数量是基本数量在某种程度上是一个选择问题。开会决定标准单位制的科学家们选择了这七种。您可能认为电流不是一个基本属性,因为它是电荷的移动速率。因此,它可以从测量电荷和时间中得出。然而,科学家们必须选择什么是基本的,他们选择了电流。这意味着电荷是通过将给定时间内通过的电流相乘得出的量。

SI UNITS 国际单位制

For each of the base units, a meeting is held every four or six years of the General Conference on Weights and Measures, under the authority of the Bureau International des Poids et Mesures in Paris. At this meeting, they either alter the definition, or agree to continue with the current definition. As we learn more and more about the universe, these definitions are gradually moving towards the fundamental constants of nature.
对于每个基本单位,在巴黎国际计量局的授权下,每四年或六年召开一次度量衡大会会议。在这次会议上,他们要么改变定义,要么同意继续当前的定义。随着我们对宇宙的了解越来越多,这些定义正逐渐朝着自然界的基本常数发展。

A fig B A standard metre, made to be exactly the length that light could travel in 1 / 299792458 1 / 299792458 1//2997924581 / 299792458 of a second.
A 图 B A 标准米,正好是光可以传播的 1 / 299792458 1 / 299792458 1//2997924581 / 299792458 一秒长度。
The current definition of each of the seven base units is listed below:
下面列出了 7 个基本单位的当前定义:
  • The kilogram is the unit of mass; it is equal to the mass of the International Prototype Kilogram, as in fig A A A\mathbf{A}.
    千克是质量单位;它等于国际原型千克的质量,如图 A A A\mathbf{A} 所示。
  • The second is the duration of 9192631770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium-133 atom.
    第二个是辐射的9192631770持续时间,对应于铯 133 原子基态的两个超精细能级之间的跃迁。
  • The metre is the length of the path travelled by light in vacuum
    米是光在真空中行进的路径长度
  • during a time interval of 1 299792458 1 299792458 (1)/(299792458)\frac{1}{299792458} of a second (see fig B).
    在一秒的 1 299792458 1 299792458 (1)/(299792458)\frac{1}{299792458} 时间间隔内(见图 B)。
  • The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 m apart in vacuum, would produce between these conductors a force equal to 2 × 10 7 2 × 10 7 2xx10^(-7)2 \times 10^{-7} newton per metre of length.
    安培是恒定电流,如果保持在两个无限长、圆形横截面可以忽略不计的平行直导体中,并在真空中相距 1 m,则在这些导体之间产生的力等于每米长度的 2 × 10 7 2 × 10 7 2xx10^(-7)2 \times 10^{-7} 牛顿。
  • The kelvin, unit of thermodynamic temperature, is the fraction 1 273.16 1 273.16 (1)/(273.16)\frac{1}{273.16} of the thermodynamic temperature of the triple point of water.
    开尔文是热力学温度的单位,是水的三相点的热力学温度的分数 1 273.16 1 273.16 (1)/(273.16)\frac{1}{273.16}
  • The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kg of carbon-12. (When the mole is used, the elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles or specified groups of such particles.)
    摩尔是一个系统的物质量,它包含的基本实体与 0.012 公斤碳 12 中的原子一样多。(使用摩尔时,必须指定基本实体,可以是原子、分子、离子、电子、其他粒子或此类粒子的指定组。

EXAM HINT 考试提示

Table A has the complete list of SI base units. You will not be asked questions about the candela in the exam.
表 A 包含国际单位制基本单位的完整列表。考试中不会询问您有关坎德拉的问题。

LEARNING TIP 学习小贴士

‘Metrology’ is the study of the science of measurement, and ‘metrics’ refers to ways of standardising measuring techniques.
“计量学”是对测量科学的研究,而“度量”是指标准化测量技术的方法。

DERIVED UNITS 派生单位

In table B you will see many of the derived units that we will study in this book, but this is only a list of those that have their own name.
在表 B 中,您将看到我们将在本书中研究的许多派生单位,但这只是那些具有自己名称的单位的列表。
 派生数量
DERIVED
QUANTITY
DERIVED QUANTITY| DERIVED | | :--- | | QUANTITY |
UNIT NAME 单元名称 UNIT SYMBOL 单位符号
 基本单位等效
BASE UNITS
EQUIVALENT
BASE UNITS EQUIVALENT| BASE UNITS | | :---: | | EQUIVALENT |
force  newton 牛顿 N kg m s 2 kg m s 2 kgms^(-2)\mathrm{kg} \mathrm{m} \mathrm{s}^{-2}
energy (work) 能量(功) joule 焦耳 J kg m 2 s 2 kg m 2 s 2 kgm^(2)s^(-2)\mathrm{kg} \mathrm{m}^{2} \mathrm{~s}^{-2}
power 权力 watt 瓦特 W W WW kg m 2 s 3 kg m 2 s 3 kgm^(2)s^(-3)\mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-3}
frequency 频率 hertz  Hz 赫兹 s 1 s 1 s^(-1)\mathrm{s}^{-1}
charge 负责 coulomb 库仑 C As 
voltage 电压 volt 伏特 V V VV kg m 2 s 3 A 1 kg m 2 s 3 A 1 kgm^(2)s^(-3)A^(-1)\mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-3} \mathrm{~A}^{-1}
resistance 电阻 ohm 欧姆 Ω Ω Omega\Omega kg m 2 s 3 A 2 kg m 2 s 3 A 2 kgm^(2)s^(-3)A^(-2)\mathrm{kg} \mathrm{m}^{2} \mathrm{~s}^{-3} \mathrm{~A}^{-2}
"DERIVED QUANTITY" UNIT NAME UNIT SYMBOL "BASE UNITS EQUIVALENT" force newton N kgms^(-2) energy (work) joule J kgm^(2)s^(-2) power watt W kgm^(2)s^(-3) frequency hertz Hz s^(-1) charge coulomb C As voltage volt V kgm^(2)s^(-3)A^(-1) resistance ohm Omega kgm^(2)s^(-3)A^(-2)| DERIVED <br> QUANTITY | UNIT NAME | UNIT SYMBOL | BASE UNITS <br> EQUIVALENT | | :--- | :---: | :---: | :---: | | force | newton | N | $\mathrm{kg} \mathrm{m} \mathrm{s}^{-2}$ | | energy (work) | joule | J | $\mathrm{kg} \mathrm{m}^{2} \mathrm{~s}^{-2}$ | | power | watt | $W$ | $\mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-3}$ | | frequency | hertz | Hz | $\mathrm{s}^{-1}$ | | charge | coulomb | C | As | | voltage | volt | $V$ | $\mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-3} \mathrm{~A}^{-1}$ | | resistance | ohm | $\Omega$ | $\mathrm{kg} \mathrm{m}^{2} \mathrm{~s}^{-3} \mathrm{~A}^{-2}$ |
table B Some well known derived units.
表 B 一些众所周知的衍生单位。

POWER PREFIXES 电源前缀

Sometimes the values we have to work with for some quantities mean that the numbers involved are extremely large or small. For example, the average distance from the Earth to the sun, measured in metres, is 150000000000 m . Scientists have made an easier system for writing such large values by adding a prefix to the unit which tells us that it has been multiplied by a very large or very small amount. In the Earth orbit example, the distance is equivalent to 150 billion metres, and the prefix giga- means multiply by a billion. So the Earth-sun distance becomes 150 gigametres, or 150 Gm .
有时,我们必须处理某些数量的值意味着所涉及的数字非常大或非常小。例如,从地球到太阳的平均距离(以米为单位)为 1500000000000 米。科学家们通过向单位添加前缀来制作一个更简单的系统来写入如此大的值,该前缀告诉我们它已经乘以非常大或非常小的量。在地球轨道示例中,距离相当于 1500 亿米,前缀 giga- 表示乘以十亿。因此,地球与太阳的距离变为 150 吉米特或 150 吉米 。
FACTOR NAME SYMBOL FACTOR NAME SYMBOL
10 1 10 1 10^(1)10^{1} deca-  da  10 1 10 1 10^(-1)10^{-1} deci- 决定- d
10 2 10 2 10^(2)10^{2} hecto- 公顷- h 10 2 10 2 10^(-2)10^{-2} centi- 厘- c
10 3 10 3 10^(3)10^{3} kilo- 公斤- k 10 3 10 3 10^(-3)10^{-3} milli- 毫- m
10 6 10 6 10^(6)10^{6} mega- 巨型- M 10 6 10 6 10^(-6)10^{-6} micro- 微- μ μ mu\mathrm{\mu}
10 9 10 9 10^(9)10^{9} giga- 吉- G 10 9 10 9 10^(-9)10^{-9} nano- 纳米 n
10 12 10 12 10^(12)10^{12} tera- 万亿 T 10 12 10 12 10^(-12)10^{-12} pico- 微微- P
10 15 10 15 10^(15)10^{15} peta- P 10 15 10 15 10^(-15)10^{-15} femto- 飞度- f
10 18 10 18 10^(18)10^{18} exa- E 10 18 10 18 10^(-18)10^{-18} atto- 阿托- a
10 21 10 21 10^(21)10^{21} zetta- 泽塔- Z 10 21 10 21 10^(-21)10^{-21} zepto- 泽普托- z
10 24 10 24 10^(24)10^{24} yotta- 约塔- Y 10 24 10 24 10^(-24)10^{-24} yocto- y
FACTOR NAME SYMBOL FACTOR NAME SYMBOL 10^(1) deca- da 10^(-1) deci- d 10^(2) hecto- h 10^(-2) centi- c 10^(3) kilo- k 10^(-3) milli- m 10^(6) mega- M 10^(-6) micro- mu 10^(9) giga- G 10^(-9) nano- n 10^(12) tera- T 10^(-12) pico- P 10^(15) peta- P 10^(-15) femto- f 10^(18) exa- E 10^(-18) atto- a 10^(21) zetta- Z 10^(-21) zepto- z 10^(24) yotta- Y 10^(-24) yocto- y| FACTOR | NAME | SYMBOL | FACTOR | NAME | SYMBOL | | :---: | :---: | :---: | :---: | :---: | :---: | | $10^{1}$ | deca- | da | $10^{-1}$ | deci- | d | | $10^{2}$ | hecto- | h | $10^{-2}$ | centi- | c | | $10^{3}$ | kilo- | k | $10^{-3}$ | milli- | m | | $10^{6}$ | mega- | M | $10^{-6}$ | micro- | $\mathrm{\mu}$ | | $10^{9}$ | giga- | G | $10^{-9}$ | nano- | n | | $10^{12}$ | tera- | T | $10^{-12}$ | pico- | P | | $10^{15}$ | peta- | P | $10^{-15}$ | femto- | f | | $10^{18}$ | exa- | E | $10^{-18}$ | atto- | a | | $10^{21}$ | zetta- | Z | $10^{-21}$ | zepto- | z | | $10^{24}$ | yotta- | Y | $10^{-24}$ | yocto- | y |
table C Prefixes used with SI units.
表 C 与 SI 单位一起使用的前缀。

CHECKPOINT 检查站

SKILLS PROBLEM SOLVING 技能 解决问题

  1. Refer to table B and answer the following questions:
    请参阅表 B 并回答以下问题:

    (a) Pick any quantity that you have studied before and explain how its base unit equivalent is shown.
    (a) 选择您之前研究过的任何数量,并解释其基本单位当量是如何显示的。

    (b) All of the derived quantity units are named after scientists. Compare their names and abbreviations. What do you notice?
    (b) 所有派生的量单位都以科学家的名字命名。比较它们的名称和缩写。您注意到了什么?
  2. Write the following in standard form:
    以标准格式编写以下内容:

    (a) 9.2 GW (a) 9.2 吉瓦
    © 6400 km © 6400 公里
    (b) 43 mm (b) 43 毫米
    (d) 44 ns. (d) 44 纳秒。
  3. Write the following using an appropriate prefix and unit symbol:
    使用适当的前缀和单位符号编写以下内容:

    (a) 3600000 joules (a) 3600000 焦耳
    © 10 millionths of an ampere
    © 10 万分之一安培

    (b) 31536000 seconds (b) 31536000 秒
    (d) 105000 hertz. (d) 105000 赫兹。

2 ESTIMATION 2 估计

LEARNING OBJECTIVES 学习目标

Estimate values for physical quantities. Use your estimates to solve problems.
估计物理量的值。使用您的估计值来解决问题。

ORDER OF MAGNITUDE 数量级

In physics, it can be very helpful to be able to make approximate estimates of values to within an order of magnitude. This means that the power of ten of your estimate is the same as the true value. For example, you are the same height as the ceiling in your classroom, if we consider the order of magnitude. The ceiling may be twice your height, but it would need to be ten times bigger to reach the next order of magnitude.
在物理学中,能够将值近似估计到一个数量级以内可能非常有帮助。这意味着您估计值的 10 的幂与真实值相同。例如,如果我们考虑数量级,您与教室中的天花板高度相同。天花板可能是你高度的两倍,但它需要大十倍才能达到下一个数量级。

This is made clearer if we express all values in standard form and then compare the power of ten. You are likely to be a thousand times taller than an ant, so we would say you are three orders of magnitude larger.
如果我们以标准形式表示所有值,然后比较 10 的幂,这会更清楚。你可能比蚂蚁高一千倍,所以我们会说你大三个数量级。

typical ant height: 1.7 mm = 1.7 × 10 3 m 1.7 mm = 1.7 × 10 3 m 1.7mm=1.7 xx10^(-3)m1.7 \mathrm{~mm}=1.7 \times 10^{-3} \mathrm{~m}
典型的蚂蚁高度: 1.7 mm = 1.7 × 10 3 m 1.7 mm = 1.7 × 10 3 m 1.7mm=1.7 xx10^(-3)m1.7 \mathrm{~mm}=1.7 \times 10^{-3} \mathrm{~m}

typical human height: 1.7 m = 1.7 × 10 m 1.7 m = 1.7 × 10 m 1.7m=1.7 xx10^(@)m1.7 \mathrm{~m}=1.7 \times 10^{\circ} \mathrm{m}
典型人体身高: 1.7 m = 1.7 × 10 m 1.7 m = 1.7 × 10 m 1.7m=1.7 xx10^(@)m1.7 \mathrm{~m}=1.7 \times 10^{\circ} \mathrm{m}


fig A We are three orders of magnitude taller than an ant.
图 A 我们比蚂蚁高三个数量级。

In many situations, physicists are not interested in specific answers, as circumstances can vary slightly and then the specific answer is incorrect. An order of magnitude answer will always be correct, unless you change the initial conditions by more than an order of magnitude. So a physicist could easily answer the question ‘What is the fastest speed of a car?’ because we don’t really want to know the exact true value. To give an exact answer would depend on knowing the model of car, and the weather and road conditions, and this answer would only be correct for that car on that day. By estimating important quantities, like a typical mass for cars, we can get an approximate - order of magnitude - answer. The reason for doing so would be that it allows us to develop ideas as possible or impossible. Also it helps us focus on developing the ideas along lines that will eventually be feasible when we get to developing a specific solution. This reduces time and money wasted by following ideas that are impossible. It also
在许多情况下,物理学家对具体的答案不感兴趣,因为情况可能会略有不同,然后具体的答案是不正确的。数量级的答案始终是正确的,除非您将初始条件更改了一个数量级以上。因此,物理学家可以很容易地回答“汽车的最快速度是多少”这个问题,因为我们真的不想知道确切的真实值。给出确切的答案将取决于了解汽车的型号、天气和路况,而这个答案仅适用于当天的那辆车。通过估计重要的量,例如汽车的典型质量,我们可以得到一个近似的 - 数量级 - 答案。这样做的原因是它允许我们尽可能或不可能地发展想法。此外,它还有助于我们专注于沿着最终在我们开发特定解决方案时可行的思路发展想法。这减少了因遵循不可能的想法而浪费的时间和金钱。它还

helps us quickly notice any miscalculations in an answer to a question. If we used an equation to calculate the answer to the fastest speed of a particular car in particular conditions, and the answer came out as 300000000 metres per second (the speed of light), we would immediately know that the answer is wrong, and re-check the calculation.
帮助我们快速注意到问题答案中的任何计算错误。如果我们用一个方程来计算特定汽车在特定条件下的最快速度的答案,答案是每秒 300000000 米(光速),我们会立即知道答案是错误的,并重新检查计算。

数量级比例
ORDER OF MAGNITUDE
SCALE
ORDER OF MAGNITUDE SCALE| ORDER OF MAGNITUDE | | :--- | | SCALE |
TYPICAL OBJECT 典型对象
1 × 10 13 m 1 × 10 13 m 1xx10^(13)m1 \times 10^{13} \mathrm{~m} size of the solar system
太阳系的大小
1 × 10 11 m 1 × 10 11 m 1xx10^(11)m1 \times 10^{11} \mathrm{~m} size of Earth's orbit around the sun
地球绕太阳轨道的大小
1 × 10 8 m 1 × 10 8 m 1xx10^(8)m1 \times 10^{8} \mathrm{~m} size of Moon's orbit around Earth
月球绕地球轨道的大小
1 × 10 4 m 1 × 10 4 m 1xx10^(4)m1 \times 10^{4} \mathrm{~m} diameter of Manchester 曼彻斯特的直径
1 × 10 0 m 1 × 10 0 m 1xx10^(0)m1 \times 10^{0} \mathrm{~m} human height 人类身高
1 × 10 3 m 1 × 10 3 m 1xx10^(-3)m1 \times 10^{-3} \mathrm{~m} ant height 蚂蚁高度
1 × 10 5 m 1 × 10 5 m 1xx10^(-5)m1 \times 10^{-5} \mathrm{~m} biological cell diameter 生物细胞直径
1 × 10 8 m 1 × 10 8 m 1xx10^(-8)m1 \times 10^{-8} \mathrm{~m} wavelength of ultraviolet light
紫外线的波长
1 × 10 10 m 1 × 10 10 m 1xx10^(-10)m1 \times 10^{-10} \mathrm{~m} diameter of an atom 原子的直径
1 × 10 14 m 1 × 10 14 m 1xx10^(-14)m1 \times 10^{-14} \mathrm{~m} diameter of an atomic nucleus
原子核的直径
"ORDER OF MAGNITUDE SCALE" TYPICAL OBJECT 1xx10^(13)m size of the solar system 1xx10^(11)m size of Earth's orbit around the sun 1xx10^(8)m size of Moon's orbit around Earth 1xx10^(4)m diameter of Manchester 1xx10^(0)m human height 1xx10^(-3)m ant height 1xx10^(-5)m biological cell diameter 1xx10^(-8)m wavelength of ultraviolet light 1xx10^(-10)m diameter of an atom 1xx10^(-14)m diameter of an atomic nucleus| ORDER OF MAGNITUDE <br> SCALE | TYPICAL OBJECT | | :--- | :--- | | $1 \times 10^{13} \mathrm{~m}$ | size of the solar system | | $1 \times 10^{11} \mathrm{~m}$ | size of Earth's orbit around the sun | | $1 \times 10^{8} \mathrm{~m}$ | size of Moon's orbit around Earth | | $1 \times 10^{4} \mathrm{~m}$ | diameter of Manchester | | $1 \times 10^{0} \mathrm{~m}$ | human height | | $1 \times 10^{-3} \mathrm{~m}$ | ant height | | $1 \times 10^{-5} \mathrm{~m}$ | biological cell diameter | | $1 \times 10^{-8} \mathrm{~m}$ | wavelength of ultraviolet light | | $1 \times 10^{-10} \mathrm{~m}$ | diameter of an atom | | $1 \times 10^{-14} \mathrm{~m}$ | diameter of an atomic nucleus |
table A Examples of object scales changing with powers of ten.
表 A 对象比例随 10 的幂而变化的示例。

FERMI QUESTIONS FERMI 问题
A fig B Enrico Fermi was one of the developers of both nuclear reactors and nuclear bombs, along with other work on particle physics, quantum physics and statistical mechanics. He was awarded the 1938 Nobel Prize for Physics for the discovery of new radioactive elements and induced radioactivity.
A 图 B 恩里科·费米 (Enrico Fermi) 是核反应堆和核弹的开发者之一,还参与了粒子物理学、量子物理学和统计力学方面的其他工作。他因发现新的放射性元素和诱导放射性而获得 1938 年诺贝尔物理学奖。

Enrico Fermi was an Italian physicist who lived from 1901 to 1954. He was a pioneer of estimation. What have become known as Fermi questions are seemingly specific questions, to which
恩里科·费米 (Enrico Fermi) 是一位意大利物理学家,生活在 1901 年至 1954 年之间。他是估算的先驱。后来被称为费米问题的东西似乎是具体的问题,其中

only an order of magnitude answer is expected. It is common for the question to appear very difficult, as we do not have enough information to work out the answer. One of Fermi’s most interesting thought experiments was a consideration of whether or not alien life exists. Over a lunch with other scientists in 1950, Fermi surprised the group by asking ‘Where is everybody?’ referring to extraterrestrials. There seems to be no evidence of the existence of alien life. That is still as true today as it was in 1950. However, when Fermi made an estimation of what would be necessary for an extraterrestrial civilisation to travel to visit us, his estimate came out at a much shorter amount of time than the age of our galaxy.
预计只有一个数量级的答案。这个问题看起来非常困难是很常见的,因为我们没有足够的信息来找出答案。费米最有趣的思想实验之一是考虑外星生命是否存在。1950 年,在与其他科学家共进午餐时,费米惊讶地问道:“大家都在哪里?似乎没有证据表明外星生命的存在。今天和 1950 年一样,这仍然适用。然而,当费米估计出外星文明访问我们所需要的条件时,他的估计结果比我们银河系的年龄要短得多。


Δ Δ Delta\Delta fig C The Fermi Paradox: even the most conservative estimates of the requirements of exploring the galaxy mean that aliens should reach Earth within ten million years of their life beginning. If they existed, they would be here.
Δ Δ Delta\Delta 图 C 费米悖论:即使是对探索银河系要求的最保守估计也意味着外星人应该在其生命开始后的一千万年内到达地球。如果他们存在,他们就会在这里。

You need to work out what steps are needed to make an estimation. First, think about what steps you would take to reach an answer, if you could have any information you wanted. Then, when the necessary data is not all available, make an estimate for the missing numbers. Making sensible assumptions is the key to solving Fermi questions.
您需要弄清楚进行估算需要哪些步骤。首先,考虑一下如果您可以获得任何您想要的信息,您将采取哪些步骤来获得答案。然后,当必要的数据并非全部可用时,对缺失的数字进行估计。做出合理的假设是解决费米问题的关键。

WORKED EXAMPLE 工作示例

Probably the most famous example of a Fermi question was this challenge to a class:
费米问题最著名的例子可能是对一个类的质询:

‘How many piano tuners are there in Chicago?’
“芝加哥有多少钢琴调音师?”

The only piece of information he provided was that the population of Chicago was 3 million.
他提供的唯一信息是芝加哥的人口为 300 万。
Step 1: How many pianos in Chicago?
第 1 步:芝加哥有多少架钢琴?

If each household is 4 people, then there are:
如果每个家庭有 4 人,那么有:

3000000 4 = 750000 3000000 4 = 750000 (3000000)/(4)=750000\frac{3000000}{4}=750000 households  3000000 4 = 750000 3000000 4 = 750000 (3000000)/(4)=750000\frac{3000000}{4}=750000 家庭
If one household in ten owns a piano, then there are:
如果十分之一的家庭拥有一架钢琴,那么就有:

750000 10 = 75000 750000 10 = 75000 (750000)/(10)=75000\frac{750000}{10}=75000 pianos  750000 10 = 75000 750000 10 = 75000 (750000)/(10)=75000\frac{750000}{10}=75000 钢琴
Step 2: How many pianos per piano tuner?
第 2 步:每个钢琴调音器有多少架钢琴?

Assume each piano needs tuning once a year. Further assume a piano tuner works 200 days a year, and can service 4 pianos a day. Each tuner can service: 200 × 4 = 800 200 × 4 = 800 200 xx4=800200 \times 4=800 pianos.
假设每架钢琴每年需要调音一次。进一步假设钢琴调音器每年工作 200 天,每天可以维修 4 架钢琴。每个调音器都可以维修: 200 × 4 = 800 200 × 4 = 800 200 xx4=800200 \times 4=800 钢琴。
Step 3: How many tuners?
第 3 步:有多少个调音器?

Each piano tuner works with 800 pianos, and there are 75000 pianos in total. So there are: 75000 800 = 94 75000 800 = 94 (75000)/(800)=94\frac{75000}{800}=94 piano tuners.
每个钢琴调音器可与 800 架钢琴配合使用,总共有 75000 架钢琴。所以有: 75000 800 = 94 75000 800 = 94 (75000)/(800)=94\frac{75000}{800}=94 钢琴调音器。

Your answer to Fermi would be ‘There are 100 piano tuners in Chicago’. This is not expected to be the exactly correct answer, but it will be correct to order of magnitude. We would not expect to find that Chicago has only 10 piano tuners, and it would be very surprising if there were 1000.
你对 Fermi 的回答是“芝加哥有 100 个钢琴调音师”。这预计不会是完全正确的答案,但它会正确到数量级。我们不会期望发现芝加哥只有 10 个钢琴调音器,如果有 1000 个,那就非常令人惊讶了。

CHECKPOINT 检查站

SKILLS ADAPTIVE LEARNING 技能自适应学习

  1. Give an order of magnitude estimate for the following quantities:
    为以下数量给出一个数量级估计值:

    (a) the height of a giraffe
    (a) 长颈鹿的身高

    (b) the mass of an apple
    (b) 苹果的质量

    © the reaction time of a human
    © 人类的反应时间

    (d) the diameter of a planet
    (d) 行星的直径

    (e) the temperature in this room.
    (e) 这个房间的温度。
  2. Answer the following Fermi questions, showing all the steps and the assumptions and estimates you make.
    回答以下 Fermi 问题,显示您所做的所有步骤以及假设和估计。

    (a) How many tennis balls would fit into a soccer stadium?
    (a) 一个足球场可以容纳多少个网球?

    (b) How many atoms are there in your body?
    (乙)你体内有多少个原子?

    © How many drops of water are there in a swimming pool?
    © 游泳池里有多少滴水?

    (d) In your lifetime, how much money will you make in total?
    (d) 在你的一生中,你总共会赚多少钱?

    (e) How many Fermi questions could Enrico Fermi have answered whilst flying from Rome to New York?
    (e) 恩里科·费米 (Enrico Fermi) 在从罗马飞往纽约的飞机上能回答多少个费米问题?

TOPIC 1 MECHANICS 主题 1:机制

1A MOTION 1A 运动

How can we calculate how fast a plane is flying, in what direction it is going and how long it will take to reach a certain destination? If you were a pilot, how would you know what force to make the engines produce and where to direct that force so your plane moves to your destination?
我们如何计算飞机的飞行速度、飞行方向以及到达某个目的地需要多长时间?如果你是一名飞行员,你怎么知道发动机产生什么力,以及将这种力引向哪里,以便你的飞机移动到目的地?

An incredible number of intricate calculations need to be done to enable a successful flight, and the basis for all of them is simple mechanics.
要想成功飞行,需要进行大量复杂的计算,而所有这些计算的基础都是简单的机械学。
This chapter explains the multiple movements of objects. It looks at how movement can be described and recorded, and then moves on to explaining why movement happens. It covers velocity and acceleration, including how to calculate these in different situations.
本章介绍对象的多种移动。它着眼于如何描述和记录运动,然后继续解释运动发生的原因。它涵盖了速度和加速度,包括如何在不同情况下计算这些值。

We only consider objects moving at speeds that could be encountered in everyday life. At these speeds (much less than the speed of light) Sir Isaac Newton succinctly described three laws of motion. With knowledge of basic geometry, we can identify aspects of movement in each dimension.
我们只考虑以日常生活中可能遇到的速度移动的物体。在这些速度(远小于光速)下,艾萨克·牛顿爵士简洁地描述了三个运动定律。有了基本几何学的知识,我们可以识别每个维度的运动方面。

Newton’s laws of motion have been constantly under test by scientists ever since he published them in 1687. Within the constraints established by Einstein in the early twentieth century, Newton’s laws have always correctly described the relationships between data collected. You may have a chance to confirm Newton’s laws in experiments of your own. With modern ICT recording of data, the reliability of such experiments is now much improved over traditional methods.
自从牛顿的运动定律于 1687 年发表以来,科学家们一直对其进行检验。在爱因斯坦在 20 世纪初建立的约束范围内,牛顿定律总是正确地描述了所收集的数据之间的关系。您可能有机会在自己的实验中证实牛顿定律。随着现代 ICT 数据的记录,此类实验的可靠性现在比传统方法大大提高。

MATHS SKILLS FOR THIS CHAPTER
本章的数学技能

  • Units of measurement (e.g. the newton, N N NN )
    测量单位(例如牛顿、 N N NN
  • Using Pythagoras’ theorem, and the angle sum of a triangle (e.g. finding a resultant vector)
    使用勾股定理和三角形的角和(例如,求结果向量)
  • Using sin, cos and tan in physical problems (e.g. resolving vectors)
    在物理问题中使用 sin、cos 和 tan(例如解析向量)
  • Using angles in regular 2D structures (e.g. interpreting force diagrams to solve problems)
    在常规 2D 结构中使用角度(例如,解释力图以解决问题)
  • Changing the subject of an equation (e.g. re-arranging the kinematics equations)
    更改方程的主题(例如,重新排列运动学方程)
  • Substituting numerical values into algebraic equations (e.g. calculating the acceleration)
    将数值代入代数方程(例如计算加速度)
  • Plotting two variables from experimental or other data, understanding that y = m x + c y = m x + c y=mx+cy=m x+c represents a linear relationship and determining the slope of a linear graph (e.g. verifying Newton’s second law experimentally)
    从实验或其他数据中绘制两个变量,理解 y = m x + c y = m x + c y=mx+cy=m x+c 代表线性关系并确定线性图的斜率(例如,通过实验验证牛顿第二定律)
  • Estimating, by graphical methods as appropriate, the area between a curve and the x x xx-axis and realising the physical significance of the area that has been determined (e.g. using a speed-time graph)
    通过适当的图形方法估计曲线和 x x xx 轴之间的面积,并实现已确定面积的物理意义(例如,使用速度-时间图)

1A 1 VELOCITY AND AGGELERATION
1A 1 速度和聚合

LEARNING OBJECTIVES 学习目标

Explain the distinction between scalar and vector quantities.
Distinguish between speed and velocity and define acceleration.
区分速度和速度并定义加速度。
  • Calculate values using equations for velocity and acceleration.
    使用速度和加速度方程计算值。


    Δ Δ Delta\Delta fig A A A\mathbf{A} These runners are accelerating to a high speed.
    Δ Δ Delta\Delta A A A\mathbf{A} 这些跑步者正在加速到高速。

LEARNING TIP 学习小贴士

The upper case Greek letter delta, Δ Δ Delta\Delta, is used in mathematics to indicate a change in a quantity. For example, Δ s Δ s Delta s\Delta s means the change in the displacement of an object, to be used here to calculate the velocity of the object.
大写希腊字母 delta Δ Δ Delta\Delta 在数学中用于表示量的变化。例如, Δ s Δ s Delta s\Delta s 表示物体位移的变化,此处用于计算物体的速度。

DID YOU KNOW? 您是否了解?

The froghopper, a 6 mm long insect, can accelerate at 4000 m s 1 4000 m s 1 4000ms^(-1)4000 \mathrm{~m} \mathrm{~s}^{-1}.
蛙蝉是一种 6 毫米长的昆虫,可以在 4000 m s 1 4000 m s 1 4000ms^(-1)4000 \mathrm{~m} \mathrm{~s}^{-1} .

LEARNING TIP 学习小贴士

Vector notation means that vectors are written in bold type to distinguish them from scalar variables.
向量表示法意味着向量以粗体书写,以区别于标量变量。
Movement is fundamental to the functioning of our universe. Whether you are running to catch a bus or want to calculate the speed required for a rocket to travel to Mars or the kinetic energy of an electron in an X-ray machine, you need to be able to work out how fast things are moving.
运动是我们宇宙运作的基础。无论您是跑步赶公共汽车,还是想计算火箭前往火星所需的速度,还是 X 光机中电子的动能,您都需要能够计算出物体的移动速度。

RATE OF MOVEMENT 移动速率

One of the simplest things we can measure is how fast an object is moving. You can calculate an object’s speed if you know the amount of time taken to move a certain distance:
我们可以测量的最简单的事情之一是物体的移动速度。如果您知道移动一定距离所花费的时间,则可以计算物体的速度:
speed ( m s 1 ) = distance ( m ) time ( s ) v = d t  speed  m s 1 =  distance  ( m ) time ( s ) v = d t {:[" speed "(ms^(-1))=(" distance "(m))/(time(s))],[v=(d)/(t)]:}\begin{aligned} \text { speed }\left(\mathrm{m} \mathrm{~s}^{-1}\right) & =\frac{\text { distance }(\mathrm{m})}{\operatorname{time}(\mathrm{s})} \\ v & =\frac{d}{t} \end{aligned}
However, the calculation for speed will only tell you how fast an object is moving. Often it is also vitally important to know in what direction this movement is taking the object. When you include the direction in the information about the rate of movement of an object, this is then known as the velocity. So, the velocity is the rate of change of displacement, where the distance in a particular direction is called the ‘displacement’.
但是,速度的计算只会告诉您物体的移动速度。通常,了解此运动将物体带向哪个方向也至关重要。当您在有关对象移动速率的信息中包含方向时,这称为速度。因此,速度是位移的变化率,其中特定方向上的距离称为“位移”。
velocity ( m s 1 ) = displacement ( m ) time ( s ) OR v = s t v = Δ s Δ t  velocity  m s 1 =  displacement  ( m )  time  ( s )  OR  v = s t v = Δ s Δ t {:[" velocity "(ms^(-1))=(" displacement "(m))/(" time "(s))],[" OR "],[v=(s)/(t)],[v=(Delta s)/(Delta t)]:}\begin{aligned} & \text { velocity }\left(\mathrm{m} \mathrm{~s}^{-1}\right)=\frac{\text { displacement }(\mathrm{m})}{\text { time }(\mathrm{s})} \\ & \text { OR } \\ & v=\frac{s}{t} \\ & v=\frac{\Delta s}{\Delta t} \end{aligned}
A fig B The displacement due north is only 75 m , whilst the actual distance this athlete has run is 300 m . So the velocity due north is much less than the actual speed.
A 图 B 正北位移只有 75 米,而这位运动员实际跑的距离是 300 米。所以正北的速度远小于实际速度。
A quantity for which the direction must be stated is known as a vector. If direction is not important, the measurement is referred to as a scalar quantity. Therefore, velocity is a vector and speed is a scalar; distance is a scalar and displacement is a vector.
必须说明方向的量称为向量。如果方向不重要,则测量值称为标量。因此,速度是一个向量,速度是一个标量;distance 是一个标量,displacement 是一个向量。
Scalar and vector quantities are not limited to measurements related to movement. Every measured quantity can be classified to include the direction (vector, e.g. force) or as being sufficiently stated by its magnitude only (scalar, e.g. mass).
标量和矢量不限于与移动相关的测量。每个测量的量都可以分类为包括方向(矢量,例如力)或仅由其大小(标量,例如质量)充分表示。

AVERAGE AND INSTANTANEOUS SPEED
平均速度和瞬时速度

In most journeys, it is unlikely that speed will remain constant throughout. As part of his training programme, an athlete in fig A wants to keep a record of his speed for all races. From rest, before the starting gun starts the race, he accelerates to a top speed. However, the race timing will be made from start to finish, and so it is most useful to calculate an average speed over the whole race. Average speed is calculated by dividing the total distance for a journey by the total time for the journey. Thus it averages out the slower and faster parts of the journey, and even includes stops.
在大多数旅程中,速度不太可能在整个过程中保持不变。作为训练计划的一部分,图 A 的一名运动员希望记录他所有比赛的速度。从休息开始,在发令枪开始之前,他加速到最高速度。但是,比赛时间将从头到尾进行,因此计算整场比赛的平均速度是最有用的。平均速度的计算方法是将旅程的总距离除以旅程的总时间。因此,它平均了旅程中较慢和较快的部分,甚至包括停靠点。

Instantaneous speed can be an important quantity, and we will look at how to measure it in the next topic.
瞬时速度可能是一个重要的量,我们将在下一个主题中研究如何测量它。

A fig C Most speed checks look at instantaneous speed, but CCTV allows police to monitor average speed over a long distance.
图 C 大多数速度检查都关注瞬时速度,但闭路电视允许警察监控远距离的平均速度。

ACCELERATION 加速度

Acceleration is defined as the rate of change of velocity. Therefore, it must include the direction in which the speed is changing, and so acceleration is a vector quantity. The equation defining acceleration is:
加速度定义为速度的变化率。因此,它必须包括速度变化的方向,因此加速度是一个矢量。定义加速度的方程式为:
acceleration ( m s 2 ) = change in velocity ( m s 1 ) time taken to change the velocity ( s )  acceleration  m s 2 =  change in velocity  m s 1  time taken to change the velocity  ( s ) " acceleration "(ms^(-2))=(" change in velocity "(ms^(-1)))/(" time taken to change the velocity "(s))\text { acceleration }\left(\mathrm{m} \mathrm{~s}^{-2}\right)=\frac{\text { change in velocity }\left(\mathrm{m} \mathrm{~s}^{-1}\right)}{\text { time taken to change the velocity }(\mathrm{s})}
OR 
a = v u t a = v u t a=(v-u)/(t)a=\frac{v-u}{t}
a = Δ v Δ t a = Δ v Δ t a=(Delta v)/(Delta t)a=\frac{\Delta v}{\Delta t}
where u u u\boldsymbol{u} is the initial velocity and v v v\boldsymbol{v} is the final velocity.
其中 u u u\boldsymbol{u} 是初始速度, v v v\boldsymbol{v} 是最终速度。

The vector nature of acceleration is very important. One of the consequences is that if an object changes only the direction of its velocity, it is accelerating, while remaining at a constant speed. Similarly, deceleration represents a negative change in velocity, and so could be stated as a negative acceleration.
加速度的矢量性质非常重要。结果之一是,如果一个物体只改变其速度的方向,它就会加速,同时保持恒定的速度。同样,减速表示速度的负变化,因此可以表示为负加速度。

EXAM HINT 考试提示

Whilst accelerations can (very briefly) be extraordinarily high, like that for the electron in question 3(b), no speed or velocity can ever be greater than the speed of light, which is 3 × 10 8 m s 1 3 × 10 8 m s 1 3xx10^(8)ms^(-1)3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}. If you calculate a speed that is higher than this, check your calculation again as it must be wrong.
虽然加速度可以(非常短暂地)非常高,就像问题 3(b) 中的电子一样,但没有速度或速度可以大于光速,即 3 × 10 8 m s 1 3 × 10 8 m s 1 3xx10^(8)ms^(-1)3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1} 。如果您计算的速度高于此速度,请再次检查您的计算,因为它一定是错误的。

CHECKPOINT 检查站

SKILLS PROBLEM SOLVING 技能 解决问题
  1. The athlete in fig B B B\mathbf{B} has taken 36 seconds from the start to reach the 300 m mark as shown. Calculate:
    如图所示,图 B B B\mathbf{B} 中的运动员从开始花了 36 秒到达 300 米大关。算:

    (a) his average speed during this 36 seconds
    (a) 他在这 36 秒内的平均速度

    (b) his average velocity due north during this 36 seconds
    (b) 他在这 36 秒内正北的平均速度

    © his average velocity due east during this 36 seconds.
    © 在这 36 秒内,他的平均速度正东。
  2. A driver in a car travelling at about 40.2 km h 1 40.2 km h 1 40.2kmh^(-1)40.2 \mathrm{~km} \mathrm{~h}^{-1} sees a cat run onto the road ahead.
    一名司机开着一辆 40.2 km h 1 40.2 km h 1 40.2kmh^(-1)40.2 \mathrm{~km} \mathrm{~h}^{-1} 车,看到一只猫跑到前面的道路上。

    (a) Convert 40.2 km h 1 40.2 km h 1 40.2kmh^(-1)40.2 \mathrm{~km} \mathrm{~h}^{-1} into a speed in m s 1 m s 1 ms^(-1)\mathrm{m} \mathrm{s}^{-1}.
    (a) 转换为 40.2 km h 1 40.2 km h 1 40.2kmh^(-1)40.2 \mathrm{~km} \mathrm{~h}^{-1} 的速度 m s 1 m s 1 ms^(-1)\mathrm{m} \mathrm{s}^{-1}

    (b) The car travels 16.5 m whilst the driver is reacting to the danger. What is his reaction time?
    (b) 当驾驶员对危险做出反应时,汽车行驶了 16.5 m。他的反应时间是什么时候?

    © The car comes to a stop in 2.5 s . What is its deceleration?
    © 汽车在 2.5 秒内停下。它的减速是多少?
  3. An electron in an X X XX-ray machine is accelerated from rest to half the speed of light in 1.7 × 10 15 s 1.7 × 10 15 s 1.7 xx10^(-15)s1.7 \times 10^{-15} \mathrm{~s}. Calculate:
    射线机中的电子在 X X XX 中从静止加速到光速的一半。 1.7 × 10 15 s 1.7 × 10 15 s 1.7 xx10^(-15)s1.7 \times 10^{-15} \mathrm{~s} 算:

    (a) the speed the electron reaches in m s 1 m s 1 ms^(-1)\mathrm{m} \mathrm{s}^{-1}
    (a) 电子达到的速度 m s 1 m s 1 ms^(-1)\mathrm{m} \mathrm{s}^{-1}

    (b) the acceleration the electron experiences.
    (b) 电子经历的加速度。

SUBJECT VOCABULARY 学科词汇

speed the rate of change of distance:
speed 距离变化的速率:
speed ( m s 1 ) = distance ( m ) time ( s ) v = d t speed m s 1 = distance ( m ) time ( s ) v = d t {:[speed(ms^(-1))=(distance(m))/(time(s))],[v=(d)/(t)]:}\begin{aligned} \operatorname{speed}\left(\mathrm{m} \mathrm{~s}^{-1}\right) & =\frac{\operatorname{distance}(\mathrm{m})}{\operatorname{time}(\mathrm{s})} \\ v & =\frac{d}{t} \end{aligned}
velocity the rate of change of displacement:
velocity 位移的变化率:
velocity ( m s 1 ) = displacement ( m ) time ( s ) v = s t OR v = Δ s Δ t  velocity  m s 1 =  displacement  ( m )  time  ( s ) v = s t  OR  v = Δ s Δ t {:[" velocity "(ms^(-1))=(" displacement "(m))/(" time "(s))],[v=(s)/(t)" OR "v=(Delta s)/(Delta t)]:}\begin{aligned} & \text { velocity }\left(\mathrm{m} \mathrm{~s}^{-1}\right)=\frac{\text { displacement }(\mathrm{m})}{\text { time }(\mathrm{s})} \\ & v=\frac{s}{t} \text { OR } v=\frac{\Delta s}{\Delta t} \end{aligned}
displacement the vector measurement of distance in a certain direction
位移 在某个方向上距离的矢量测量

vector a quantity that must have both magnitude and direction scalar a quantity that has magnitude only average speed speed for a whole journey, calculated by dividing the total distance for a journey by the total time for the journey:
vector 一个必须同时具有 magnitude 和 direction scalar 的量 一个只有 magnitude 的量 整个旅程的平均速度 速度,计算方法是将旅程的总距离除以旅程的总时间:
average speed ( m s 1 ) = total distance ( m ) total time ( s )  average speed  m s 1 =  total distance  ( m )  total time  ( s ) " average speed "(ms^(-1))=(" total distance "(m))/(" total time "(s))\text { average speed }\left(\mathrm{m} \mathrm{~s}^{-1}\right)=\frac{\text { total distance }(\mathrm{m})}{\text { total time }(\mathrm{s})}
instantaneous speed the speed at any particular instant in time on a journey, which can be found from the gradient of the tangent to a distance-time graph at that time acceleration the vector defined as the rate of change of velocity:
Instantaneous speed 行程中任何特定时刻的速度,可以从该时间的距离-时间图的切线梯度中找到加速度向量定义为速度变化率:
acceleration ( m s 2 ) = change in velocity ( m s 1 ) time taken to change the velocity ( s )  acceleration  m s 2 =  change in velocity  m s 1  time taken to change the velocity  ( s ) " acceleration "(ms^(-2))=(" change in velocity "(ms^(-1)))/(" time taken to change the velocity "(s))\text { acceleration }\left(\mathrm{m} \mathrm{~s}^{-2}\right)=\frac{\text { change in velocity }\left(\mathrm{m} \mathrm{~s}^{-1}\right)}{\text { time taken to change the velocity }(\mathrm{s})}
a = v u t OR a = Δ v Δ t a = v u t  OR  a = Δ v Δ t a=(v-u)/(t)" OR "a=(Delta v)/(Delta t)a=\frac{v-u}{t} \text { OR } a=\frac{\Delta v}{\Delta t}

LEARNING OBJECTIVES 学习目标

  • Interpret displacement-time graphs, velocity-time graphs and acceleration-time graphs.
    解释位移时间图、速度时间图和加速时间图。
  • Make calculations from these graphs.
    根据这些图表进行计算。
  • Understand the graphical representations of accelerated motion.
    了解加速运动的图形表示。
One of the best ways to understand the movements of an object whilst on a journey is to plot a graph of the position of the object over time. Such a graph is known as a displacement-time graph. A velocity-time graph will also provide detail about the movements involved. A velocity-time graph can be produced from direct measurements of the velocity or generated from calculations made using the displacement-time graph.
了解物体在旅途中运动的最佳方法之一是绘制物体位置随时间变化的图表。这样的图形称为位移-时间图。速度-时间图还将提供有关所涉及的运动的详细信息。速度-时间图可以通过直接测量速度来生成,也可以从使用位移-时间图进行的计算中生成。

DISPLACEMENT-TIME GRAPHS 位移时间图

If we imagine a boat trip on a river, we could monitor the location of the boat over the hour that it travels for and plot the displacementtime graph for these movements. Depending on what information we want the graph to provide, it is often simpler to draw a distancetime graph in which the direction of movement is ignored.
如果我们想象在河流上乘船旅行,我们可以在船行驶的一小时内监控船的位置,并绘制这些运动的位移时间图。根据我们希望图形提供的信息,绘制忽略移动方向的 distancetime 图形通常更简单。

The graphs shown in fig A are examples of plotting position against time, and show how a distance-time graph cannot decrease with time. A displacement-time graph could have parts of it in the negative portions of the y y yy-axis, if the movement went in the opposite direction at some points in time.
图 A 中显示的图表是绘制位置与时间的关系示例,并显示了距离-时间图如何不能随时间而减小。如果运动在某些时间点朝相反的方向发展,则位移时间图的一部分可能位于 -轴的 y y yy 负部分。

The simplest thing we could find from these graphs is how far an object has moved in a certain time. For example, in fig A, both the graphs show that in the first 15 minutes the boat moved 150 m . Looking at the time from 40 to 48 minutes, both show that the boat travelled 120 m , but the displacement-time graph is in the negative region of the y y yy-axis, showing the boat was moving down river from the starting point - the opposite direction to the places it had been in the first 40 minutes.
从这些图表中,我们可以找到最简单的信息是物体在一定时间内移动了多远。例如,在图 A 中,两张图都显示船在前 15 分钟内移动了 150 m。查看 40 到 48 分钟的时间,两者都显示船行驶了 120 m,但位移时间图位于 y y yy 轴的负区域,显示船从起点顺流而下 - 与前 40 分钟的位置相反。
During the period from 20 to 25 minutes, both graphs have a flat line at a constant value, that shows no change in the distance or displacement. This means the boat was not moving - a flat line on a distance-time ( d t ) ( d t ) (d-t)(d-t) graph means the object is stationary. From 20 to 25 minutes on the velocity-time ( v t ) ( v t ) (v-t)(\boldsymbol{v}-t) graph of this journey (see fig B) the line would be at a velocity of 0 m s 1 0 m s 1 0ms^(-1)0 \mathrm{~m} \mathrm{~s}^{-1}.
在 20 到 25 分钟的时间内,两张图都有一条恒定值的平线,表明距离或位移没有变化。这意味着船没有移动 - 距离-时间 ( d t ) ( d t ) (d-t)(d-t) 图上的平线表示物体是静止的。在这次旅程的速度-时间 ( v t ) ( v t ) (v-t)(\boldsymbol{v}-t) 图上(见图 B),从 20 到 25 分钟,该线的速度为 0 m s 1 0 m s 1 0ms^(-1)0 \mathrm{~m} \mathrm{~s}^{-1}

SPEED AND VELOCITY FROM d t d t d-td-t GRAPHS
SPEED 和 VELOCITY from d t d t d-td-t Graphs

The gradient of the d t d t d-td-t graphs in fig A A A\mathbf{A} will tell us how fast the boat was moving. Gradient is found from the ratio of changes in the y y yy-axis divided by the corresponding change on the x x xx-axis, so:
图中 d t d t d-td-t 图表的梯度 A A A\mathbf{A} 将告诉我们船的移动速度有多快。梯度是通过 y y yy -轴上的变化除以 x x xx -轴上的相应变化的比率得出的,因此:

for a distance-time graph:
对于距离-时间图:
gradient = distance ( m ) time ( s ) = speed ( m s 1 ) v = d t  gradient  = distance ( m ) time ( s ) = speed m s 1 v = d t {:[" gradient "=(distance(m))/(time(s))=speed(ms^(-1))],[v=(d)/(t)]:}\begin{aligned} \text { gradient } & =\frac{\operatorname{distance}(\mathrm{m})}{\operatorname{time}(\mathrm{s})}=\operatorname{speed}\left(\mathrm{m} \mathrm{~s}^{-1}\right) \\ v & =\frac{d}{t} \end{aligned}
for a displacement-time graph:
对于 Displacement-Time 图表:
gradient = displacement ( m ) time ( s ) = velocity ( m s 1 ) v = Δ s Δ t  gradient  =  displacement  ( m )  time  ( s ) =  velocity  m s 1 v = Δ s Δ t {:[" gradient "=(" displacement "(m))/(" time "(s))=" velocity "(ms^(-1))],[v=(Delta s)/(Delta t)]:}\begin{aligned} \text { gradient } & =\frac{\text { displacement }(\mathrm{m})}{\text { time }(\mathrm{s})}=\text { velocity }\left(\mathrm{m} \mathrm{~s}^{-1}\right) \\ v & =\frac{\Delta s}{\Delta t} \end{aligned}fig A A comparison of the displacement-time graph of the boat trip up and down a river with its corresponding distance-time graph.
图 A 乘船在河流上下游的位移-时间图与其相应的距离-时间图的比较。
For example, the first 15 minutes of the boat trip in fig A represents a time of 900 seconds. In this time, the boat travelled 150 m . Its velocity is:
例如,图 A 中乘船旅行的前 15 分钟表示 900 秒的时间。在这段时间里,船行驶了 150 m。它的速度为:
v = Δ s Δ t = 150 900 = 0.167 m s 1 up river v = Δ s Δ t = 150 900 = 0.167 m s 1  up river  v=(Delta s)/(Delta t)=(150)/(900)=0.167ms^(-1)" up river "v=\frac{\Delta s}{\Delta t}=\frac{150}{900}=0.167 \mathrm{~m} \mathrm{~s}^{-1} \text { up river }

VELOCITY-TIME GRAPHS 速度-时间图

A velocity-time graph will show the velocity of an object over time. We calculated that the velocity of the boat on the river was 0.167 m s 1 0.167 m s 1 0.167ms^(-1)0.167 \mathrm{~m} \mathrm{~s}^{-1} up river for the first 15 minutes of the journey. Looking at the graph in fig B, you can see that the line is constant at + 0.167 m s 1 + 0.167 m s 1 +0.167ms^(-1)+0.167 \mathrm{~m} \mathrm{~s}^{-1} for the first 15 minutes.
速度-时间图将显示物体随时间变化的速度。我们计算出,在旅程的前 15 分钟内,河上的船速是 0.167 m s 1 0.167 m s 1 0.167ms^(-1)0.167 \mathrm{~m} \mathrm{~s}^{-1} 逆流而上。查看图 B 中的图表,您可以看到该线 + 0.167 m s 1 + 0.167 m s 1 +0.167ms^(-1)+0.167 \mathrm{~m} \mathrm{~s}^{-1} 在前 15 分钟内保持不变。

Also notice that the velocity axis includes negative values, so that the difference between travelling up river (positive y y yy-axis values) and down river (negative y y yy-axis values) can be represented.
另请注意,速度轴包含负值,因此可以表示逆流而上(正 y y yy 轴值)和顺流而下(负 y y yy 轴值)之间的差异。


fig B 图 B
Velocity-time graph of the boat trip.
乘船行程的速度-时间图。

ACCELERATION FROM v t v t v-tv-t GRAPHS
图形加速度 v t v t v-tv-t

Acceleration is defined as the rate of change in velocity. In order to calculate the gradient of the line on a v t v t v-t\boldsymbol{v}-t graph, we must divide a change in velocity by the corresponding time difference. This exactly matches with the equation for acceleration:
加速度定义为速度的变化率。为了计算 v t v t v-t\boldsymbol{v}-t 图形上线条的梯度,我们必须将速度变化除以相应的时间差。这与加速度方程完全匹配:
gradient = Δ v Δ t = v u t = acceleration  gradient  = Δ v Δ t = v u t =  acceleration  " gradient "=(Delta v)/(Delta t)=(v-u)/(t)=" acceleration "\text { gradient }=\frac{\Delta v}{\Delta t}=\frac{v-u}{t}=\text { acceleration }
For example, between 15 and 20 minutes on the graphs, the boat slows evenly to a stop. The acceleration here can be calculated as the gradient:
例如,在图表上的 15 到 20 分钟之间,船会均匀地减速直至停止。这里的加速度可以计算为梯度:
gradient = Δ v Δ t = v u t = 0 0.167 5 × 60 = 0.167 300 = 0.0006 m s 2  gradient  = Δ v Δ t = v u t = 0 0.167 5 × 60 = 0.167 300 = 0.0006 m s 2 " gradient "=(Delta v)/(Delta t)=(v-u)/(t)=(0-0.167)/(5xx60)=(-0.167)/(300)=-0.0006ms^(-2)\text { gradient }=\frac{\Delta v}{\Delta t}=\frac{v-u}{t}=\frac{0-0.167}{5 \times 60}=\frac{-0.167}{300}=-0.0006 \mathrm{~m} \mathrm{~s}^{-2}
So the acceleration is: a = 0.6 × 10 3 m s 2 a = 0.6 × 10 3 m s 2 a=-0.6 xx10^(-3)ms^(-2)\boldsymbol{a}=-0.6 \times 10^{-3} \mathrm{~m} \mathrm{~s}^{-2}.
所以加速度为: a = 0.6 × 10 3 m s 2 a = 0.6 × 10 3 m s 2 a=-0.6 xx10^(-3)ms^(-2)\boldsymbol{a}=-0.6 \times 10^{-3} \mathrm{~m} \mathrm{~s}^{-2}

DISTANCE TRAVELLED FROM v t v t v-tv-t GRAPHS
v t v t v-tv-t 图形中移动的距离

Speed is defined as the rate of change in distance:
速度定义为距离的变化率:
v = d t d = v × t v = d t d = v × t {:[v=(d)/(t)],[:.quad d=v xx t]:}\begin{aligned} & v=\frac{d}{t} \\ & \therefore \quad d=v \times t \end{aligned}
As the axes on the v t v t v-t\boldsymbol{v}-t graph represent velocity and time, an area on the graph represents the multiplication of velocity × × xx\times time, which gives distance. So to find the distance travelled from a v t v t v-t\boldsymbol{v}-t graph, find the area between the line and the x x xx-axis.
由于 v t v t v-t\boldsymbol{v}-t 图形上的轴表示速度和时间,因此图形上的区域表示速度 × × xx\times 时间的乘积,从而得出距离。因此,要找到从 v t v t v-t\boldsymbol{v}-t 图表出发的距离,请找到线和 x x xx -轴之间的区域。


Δ Δ Delta\Delta fig C In the first 15 minutes ( 900 seconds) the distance travelled by the boat moving at 0.167 m s 1 0.167 m s 1 0.167ms^(-1)0.167 \mathrm{~m} \mathrm{~s}^{-1} is given by the area between the line and the x x xx-axis: d = v × t = 0.167 × 900 = 150 m d = v × t = 0.167 × 900 = 150 m d=v xx t=0.167 xx900=150md=\boldsymbol{v} \times t=0.167 \times 900=150 \mathrm{~m}.
Δ Δ Delta\Delta 图 C 在前 15 分钟(900 秒)中,船只行驶的距离 0.167 m s 1 0.167 m s 1 0.167ms^(-1)0.167 \mathrm{~m} \mathrm{~s}^{-1} 由线和 x x xx 轴之间的面积给出: d = v × t = 0.167 × 900 = 150 m d = v × t = 0.167 × 900 = 150 m d=v xx t=0.167 xx900=150md=\boldsymbol{v} \times t=0.167 \times 900=150 \mathrm{~m}
If we are only interested in finding the distance moved, this also works for a negative velocity. You find the area from the line up to the time axis. This idea will still work for a changing velocity. Find the area under the line and you have found the distance travelled. For example, from 0 to 20 minutes, the area under the line, all the way down to the x x xx-axis, is a trapezium, so we need to find that area. To calculate the whole distance travelled in the journey for the first 40 minutes, we would have to find the areas under the four separate stages ( 0 15 0 15 0-150-15 minutes; 15 20 15 20 15-2015-20 minutes; 20 25 20 25 20-2520-25 minutes; and 25-40 minutes) and then add these four answers together.
如果我们只对找到移动的距离感兴趣,这也适用于负速度。您可以找到从线到时间轴的区域。这个想法仍然适用于不断变化的速度。找到线下方的区域,您就找到了行驶的距离。例如,从 0 到 20 分钟,线下方的区域,一直到 x x xx -轴,是一个梯形,所以我们需要找到那个区域。要计算前 40 分钟旅程中行驶的总距离,我们必须找到四个不同阶段( 0 15 0 15 0-150-15 分钟; 15 20 15 20 15-2015-20 纪要; 20 25 20 25 20-2520-25 纪要;和 25-40 分钟),然后将这四个答案相加。

PRACTICAL SKILLS 实践技能

Finding the acceleration due to gravity by multiflash photography
通过多重闪光摄影求重力加速度

Using a multiflash photography technique, or a video recording that can be played back frame by frame, we can observe the falling motion of a small object such as a marble (see fig D). We need to know the time between frames.
使用多闪摄影技术,或可以逐帧播放的视频记录,我们可以观察弹珠等小物体的下落运动(见图 D)。我们需要知道帧之间的时间。

From each image of the falling object, measure the distance it has fallen from the scale in the picture A carefully drawn distance-time graph will show a curve as the object accelerates. From this curve, take regular measurements of the gradient by drawing tangents to the curve. These gradients show the instantaneous speed at each point on the curve.
从下落物体的每张图像中,测量它与图片中刻度的距离 精心绘制的距离-时间图将在物体加速时显示曲线。从此曲线中,通过绘制曲线的切线来定期测量渐变。这些梯度显示曲线上每个点的瞬时速度。


fig D Multiflash photography allows us to capture the accelerating movement of an object falling under gravity.
图 D Multiflash 摄影使我们能够捕捉到物体在重力作用下落的加速运动。
Plotting these speeds on a velocity-time graph should show a straight line, as the acceleration due to gravity is a constant value. The gradient of the line on this v v v\boldsymbol{v} - t t tt graph will be the acceleration due to gravity, g g gg.
在速度-时间图上绘制这些速度应该显示一条直线,因为重力引起的加速度是一个常数值。此 v v v\boldsymbol{v} - t t tt 图上线条的梯度将是重力加速度 g g gg
Safety Note: Persons with medical conditions such as epilepsy or migraine may be adversely affected by multiflash photography.
安全提示:患有癫痫或偏头痛等疾病的人可能会受到多重闪光摄影的不利影响。

ACCELERATION-TIME GRAPHS 加速时间图

Acceleration-time graphs show how the acceleration of an object changes over time. In many instances the acceleration is zero or a constant value, in which case an acceleration-time ( a t ) ( a t ) (a-t)(\boldsymbol{a}-t) graph is likely to be of relatively little interest. For example, the object falling in our investigation above will be accelerated by gravity throughout. Assuming it is relatively small, air resistance will be minimal, and the a t a t a-t\boldsymbol{a}-t graph of its motion would be a horizontal line at a = 9.81 ms 2 a = 9.81 ms 2 a=-9.81ms^(-2)\boldsymbol{a}=-9.81 \mathrm{~ms}^{-2}. Compare this with your results to see how realistic it is to ignore air resistance.
加速时间图表显示对象的加速度如何随时间变化。在许多情况下,加速度为零或恒定值,在这种情况下,加速度时间 ( a t ) ( a t ) (a-t)(\boldsymbol{a}-t) 图可能相对不感兴趣。例如,我们上面调查中坠落的物体将在整个过程中受到重力的加速。假设它相对较小,空气阻力将最小,其运动 a t a t a-t\boldsymbol{a}-t 图将是一条位于 的 a = 9.81 ms 2 a = 9.81 ms 2 a=-9.81ms^(-2)\boldsymbol{a}=-9.81 \mathrm{~ms}^{-2} 水平线。将此结果与您的结果进行比较,以了解忽略空气阻力的现实程度。
For a larger object falling for a long period, such as a skydiver, then the acceleration will change over time as the air resistance increases with speed.
对于长时间下落的较大物体,例如跳伞者,那么加速度会随着空气阻力的增加而随时间而变化。


fig E Acceleration-time graph for a skydiver.
图 E 跳伞运动员的加速时间图。
The weight of a skydiver is constant, so the resultant force will be decreasing throughout, which means that the acceleration will also reduce (see Section 1A.5). The curve would look like that in fig E. See Section 2A. 4 for more details on falling objects and terminal velocity.
跳伞运动员的重量是恒定的,因此合力将始终减小,这意味着加速度也会减小(参见第 1A.5 节)。曲线如图 E 所示。有关下落物体和终端速度的更多详细信息,请参见第 2A. 4 节。

EXAM HINT 考试提示

Remember that the gradient of a distance-time graph represents speed or velocity. So if the line is curved, the changing gradient indicates a changing speed, which you can describe as the same as the changes in gradient.
请记住,距离-时间图的梯度表示速度或速度。因此,如果线条是弯曲的,则变化的渐变表示变化的速度,您可以将其描述为与渐变的变化相同。

CHECKPOINT 检查站

SKILLS 技能
INTERPRETATION 解释
(OF GRAPHICAL DATA) (图形数据)
  1. Describe in as much detail as you can, including calculated values, what happens in the bicycle journey shown on the d t d t d-td-t graph in fig F F F\mathbf{F}.
    尽可能详细地描述,包括计算值,图中 d t d t d-td-t 图表所示的自行车旅程中发生的情况 F F F\mathbf{F}


    fig F F F\mathbf{F} Distance-time graph of a bike journey.
    F F F\mathbf{F} 图 自行车行程的距离-时间图。
  2. Describe in as much detail as you can, including calculated values, what happens in the car journey shown on the v v v\boldsymbol{v} - t t tt graph in fig G G GG.
    尽可能详细地描述,包括计算值,图 v v v\boldsymbol{v} - t t tt G G GG 中所示的汽车旅程中发生的情况。


    fig G Velocity-time graph of a car journey.
    图 G 汽车行程的速度-时间图。
  3. From fig B, calculate the distance travelled by the boat from 40 to 60 minutes.
    根据图 B,计算船只行驶 40 到 60 分钟的距离。

SUBJECT VOCABULARY 学科词汇

displacement-time graph a graph showing the positions visited on a journey, with displacement on the y y yy-axis and time on the x x xx-axis. velocity-time graph a graph showing the velocities on a journey, with velocity on the y y yy-axis and time on the x x xx-axis.
displacement-time graph (置换时间图) 显示历程中访问的位置的图表,其中 y y yy displacement 在 -轴上,time 在 x x xx -轴上。Velocity-Time Graph (速度-时间) 显示旅程中速度的图表,其中 y y yy Speed 在 -轴上,Time 在 x x xx -轴上。

gradient the slope of a line or surface
渐变线或表面的坡度

LEARNING OBJECTIVES
- Add two or more vectors by drawing.
- Add two perpendicular vectors by calculation.
学习目标 - 通过绘制添加两个或多个矢量。 - 通过计算添加两个垂直向量。

Forces are vectors. This means that measuring their magnitude is important, but equally important is knowing the direction in which they act. In order to calculate the overall effect of multiple forces acting on the same object, we can use vector addition to work out the resultant force. This resultant force can be considered as a single force that has the same effect as all the individual forces combined.
力是矢量。这意味着测量它们的大小很重要,但同样重要的是了解它们的作用方向。为了计算作用在同一物体上的多个力的整体效果,我们可以使用矢量加法来计算合力。这个合力可以看作是一个单一的力,其效果与所有单个力的总和相同。

ADDING FORCES IN THE SAME LINE
在同一条直线上添加力

If two or more forces are acting along the same line, then combining them is simply a case of adding or subtracting their magnitudes depending on their directions.
如果两个或多个力沿同一条线作用,那么将它们组合起来只是根据它们的方向增加或减少它们的大小。


Δ f i g A Δ f i g A Delta figA\Delta \boldsymbol{f i g} \mathbf{A} Adding forces in the same line requires a consideration of their comparative directions.
Δ f i g A Δ f i g A Delta figA\Delta \boldsymbol{f i g} \mathbf{A} 在同一条线上添加力需要考虑它们的比较方向。

ADDING PERPENDICULAR FORCES
添加垂直力

The effect on an object of two forces that are acting at right angles (perpendicular) to each other will be the vector sum of their individual effects. We need to add the sizes with consideration for the directions in order to find the resultant.
两个彼此成直角(垂直)作用的力对物体的影响将是它们各自效应的向量和。我们需要在考虑方向的情况下添加大小,以便找到结果。

A fig B These two rugby players are each putting a force on their opponent. The forces are at right angles, so the overall effect would be to move him in a third direction, which we could calculate.
A 图 B 这两名橄榄球运动员各自对他们的对手施加了压力。力是直角的,所以整体效果是让他向第三个方向移动,我们可以计算出来。

MAGNITUDE OF THE RESULTANT FORCE
合力的大小

To calculate the resultant magnitude of two perpendicular forces, we can draw them, one after the other, as the two sides of a right-angled triangle and use Pythagoras’ theorem to calculate the size of the hypotenuse.
为了计算两个垂直力的合力大小,我们可以将它们一个接一个地绘制为直角三角形的两条边,并使用毕达哥拉斯定理计算斜边的大小。


Δ Δ Delta\Delta fig C C C\mathbf{C} The resultant force here is calculated using Pythagoras’ theorem: F = ( 70 2 + 110 2 ) = 130 N F = 70 2 + 110 2 = 130 N F=sqrt((70^(2)+110^(2)))=130NF=\sqrt{\left(70^{2}+110^{2}\right)}=130 \mathrm{~N}
Δ Δ Delta\Delta C C C\mathbf{C} 图 这里的合力是使用勾股定理计算的: F = ( 70 2 + 110 2 ) = 130 N F = 70 2 + 110 2 = 130 N F=sqrt((70^(2)+110^(2)))=130NF=\sqrt{\left(70^{2}+110^{2}\right)}=130 \mathrm{~N}

DIRECTION OF THE RESULTANT FORCE
合力的方向

As forces are vectors, when we find a resultant force it must have both magnitude and direction. For perpendicular forces (vectors), trigonometry will determine the direction.
由于力是矢量,当我们找到合力时,它必须同时具有大小和方向。对于垂直力(矢量),三角学将确定方向。


Δ Δ Delta\Delta fig D D D\mathbf{D} The resultant force here is at an angle up from the horizontal of: θ = tan 1 ( 70 110 ) = 32 θ = tan 1 70 110 = 32 theta=tan^(-1)((70)/(110))=32^(@)\theta=\tan ^{-1}\left(\frac{70}{110}\right)=32^{\circ}
Δ Δ Delta\Delta D D D\mathbf{D} 图 这里的合力与以下的水平线成一定角度: θ = tan 1 ( 70 110 ) = 32 θ = tan 1 70 110 = 32 theta=tan^(-1)((70)/(110))=32^(@)\theta=\tan ^{-1}\left(\frac{70}{110}\right)=32^{\circ}

EXAM HINT 考试提示

Always take care to state where the angle for a vector’s direction is measured. For example, in fig D D DD, the angle should be stated as 32 32 32^(@)32^{\circ} up from the horizontal. This is most easily expressed on a diagram of the situation, where you draw in the angle.
始终注意说明矢量方向的角度的测量位置。例如,在图 D D DD 中,角度应表示 32 32 32^(@)32^{\circ} 为从水平方向向上。这在情况图上最容易表达,您可以在其中绘制角度。

ADDING TWO NON-PERPENDICULAR FORCES
添加两个非垂直力

The geometry of perpendicular vectors makes the calculation of the resultant simple. We can find the resultant of any two vectors by drawing one after the other, and then the resultant will be the third side of the triangle from the start of the first one to the end of the second one. A scale drawing of the vector triangle will allow measurement of the size and direction of the resultant.
垂直向量的几何结构使结果的计算变得简单。我们可以通过一个接一个地绘制来找到任意两个向量的结果,然后结果将是三角形的第三条边,从第一个向量的起点到第二个向量的终点。矢量三角形的比例图将允许测量结果的大小和方向。

A fig E The resultant force here can be found by scale drawing the two forces, and then measurement of the resultant on the drawing using a ruler and a protractor.
A 图 E 这里的合力可以通过按比例绘制两个力,然后使用尺子和量角器在图纸上测量合力来找到。

70 110 N 70 110 N 70 obrace(110(N))70 \overbrace{110 \mathrm{~N}}

A fig G Free-body force diagram of a rugby player (red circle). The forces from the tacklers are marked on as force arrows.
图 G 橄榄球运动员的自由体力图(红色圆圈)。铲球手的力量被标记为力箭。

THE PARALLELOGRAM RULE 平行四边形规则

There is another method for finding the resultant of two non-perpendicular forces (or vectors) by scale drawing, which can be easier to use. This is called the parallelogram rule. Draw the two vectors to scale - at the correct angle and scaled so their length represents the magnitude - starting from the same point. Then draw the same two vectors again parallel to the original ones, so that they form a parallelogram, as shown in fig F F F\mathbf{F}. The resultant force (or vector) will be the diagonal across the parallelogram from the starting point.
还有另一种方法可以通过比例绘图找到两个非垂直力(或矢量)的合力,这种方法更容易使用。这称为平行四边形规则。绘制两个要缩放的向量 - 以正确的角度并缩放,使它们的长度代表大小 - 从同一点开始。然后再次绘制相同的两个向量,使其平行于原始向量,使它们形成一个平行四边形,如图 F F F\mathbf{F} 所示。合力(或矢量)将是从起点穿过平行四边形的对角线。


fig F Finding the resultant vector using the parallelogram rule.
图 F 使用平行四边形规则求结果向量。

EXAM HINT 考试提示

The vector addition rules shown on these pages work for all vectors, not just forces. They are useful only for co-planar vectors, which means vectors that are in the same plane. If we have more than two vectors that are in more than one plane, add two vectors together first, in their plane, and then add the resultant to the next vector using these rules again. Keep doing this until all the vectors have been added in.
这些页面上显示的向量添加规则适用于所有向量,而不仅仅是力。它们仅对共面向量有用,共面向量是指位于同一平面中的向量。如果我们有两个以上的向量在多个平面中,请先在它们的平面上将两个向量相加,然后再次使用这些规则将结果添加到下一个向量中。继续执行此操作,直到添加完所有向量。

FREE-BODY FORCE DIAGRAMS 自由体力图

If we clarify what forces are acting on an object, it can be simpler to calculate how it will move. To do this, we usually draw a free-body force diagram, which has the object isolated, and all the forces that act on it drawn in at the points where they act. Forces acting on other objects, and those other objects, are not drawn. For example, fig G could be said to be a free-body force diagram of the rugby player being tackled in fig B, and this would lead us to draw fig C C C\mathbf{C} and fig D D D\mathbf{D} to make our resultant calculations.
如果我们阐明作用在物体上的力,那么计算它将如何移动就会更简单。为此,我们通常绘制一个自由体力图,该图将物体隔离开来,并且作用在它上的所有力都集中在它们作用的点上。不会绘制作用在其他对象和这些其他对象上的力。例如,图 G 可以说是图 B 中橄榄球运动员被铲球的自由体力图,这将导致我们绘制 fig C C C\mathbf{C} 和 fig D D D\mathbf{D} 来进行结果计算。

CHECKPONT

  1. Work out the resultant force on a toy car if it has the following forces acting on it:
    如果玩具车上有以下力,请计算玩具车上的合力:
  • rubber band motor driving forwards 8.4 N
    橡皮筋电机向前驱动 8.4 N
  • air resistance 0.5 N 空气阻力 0.5 N
  • friction 5.8 N 摩擦力 5.8 N
  • child’s hand pushing forward 10 N .
    孩子的手向前推 10 N .
  1. As a small plane accelerates to take off, the lift force on it is 6000 N vertically upwards, whilst the thrust is 2800 N horizontally forwards. What is the resultant of these forces on the plane?
    当一架小型飞机加速起飞时,其垂直向上的升力为 6000 N,而水平向前的推力为 2800 N。这些力在飞机上的合力是什么?
  2. Draw a free-body force diagram of yourself sitting on your chair.
    画一张你自己坐在椅子上的自由体力图。
  3. (a) Draw the scale diagram of fig E, and work out what the resultant force would be.
    (a) 画出图 E 的比例图,并计算出合力是多少。

    (b) Use the parallelogram rule, as in fig F F F\mathbf{F}, to check your answer to part (a).
    (b) 使用平行四边形规则,如图 F F F\mathbf{F} ,检查您对 (a) 部分的回答。
  4. In order to try and recover a car stuck in a muddy field, two tractors pull on it. The first acts at an angle of 20 20 20^(@)20^{\circ} left of the forwards direction with a force of 2250 N . The second acts 15 15 15^(@)15^{\circ} to the right of the forwards direction with a force of 2000 N . Draw a scale diagram of the situation and find the resultant force on the stuck car.
    为了试图找回一辆被困在泥泞田野中的汽车,两台拖拉机拉着它。第一个以 2250 N 的力以向前方向 20 20 20^(@)20^{\circ} 的左侧角度作用。第二个作用 15 15 15^(@)15^{\circ} 在前进方向的右侧,力为 2000 N 。绘制情况的比例图并找到卡住的汽车上的合力。

SUBJECT VOCABULARY 学科词汇

resultant force the total force (vector sum) acting on a body when all the forces are added together accounting for their directions
合力 当所有力相加时作用在物体上的总力(矢量和),考虑它们的方向

free-body force diagram diagram showing an object isolated, and all the forces that act on it are drawn in at the points where they act, using arrows to represent the forces
自由体力图图显示了一个孤立的物体,作用在它上的所有力都被吸引到它们作用的点上,使用箭头表示力

LEARNING OBJECTIVES 学习目标

  • Calculate the moment of a force.Apply the principle of moments.
    计算力矩。应用力矩原理。

    Find the centre of gravity of an object.
    找到对象的重心。

    Forces on an object could act so that the object does not start to move along, but instead rotates about a fixed pivot. If the object is fixed so that it cannot rotate, it will bend.
    对象上的力可以起作用,使对象不会开始移动,而是围绕固定轴旋转。如果对象被固定到无法旋转,它将弯曲。

THE MOMENT OF A FORCE
力矩

A fig A A force acts on a beam fixed at a point. The moment of a force causes rotation or, in this case, bending.
图 A A 力作用在固定在某一点的横梁上。力矩会导致旋转,或者在这种情况下导致弯曲。

The tendency to cause rotation is called the moment of a force. It is calculated from:
引起旋转的趋势称为力矩。计算公式为:

moment ( Nm ) = ( Nm ) = (Nm)=(\mathrm{Nm})= force ( N ) × ( N ) × (N)xx(\mathrm{N}) \times perpendicular distance from the pivot to the line of action of the force ( m ) ( m ) (m)(m)
力矩 ( Nm ) = ( Nm ) = (Nm)=(\mathrm{Nm})= ( N ) × ( N ) × (N)xx(\mathrm{N}) \times 从枢轴到力 ( m ) ( m ) (m)(m) 的作用线的垂直距离

moment = F x = F x =Fx=\boldsymbol{F} x 时刻 = F x = F x =Fx=\boldsymbol{F} x
A fig B The calculation of moment only considers the perpendicular distance between the line of action of the force and the axis of rotation, through the pivot point. When free to rotate, a body will turn in the direction of any net moment.
A 图 B 力矩的计算仅考虑力的作用线与旋转轴之间的垂直距离,通过枢轴点。当自由旋转时,物体将沿任何净力矩的方向转动。

PRINCIPLE OF MOMENTS 矩量原理


fig C Balanced moments create an equilibrium situation.
图 C 平衡矩产生平衡情况。

If we add up all the forces acting on an object and the resultant force, accounting for their directions, is zero, then the object will be in equilibrium. Therefore it will remain stationary or, if it is already moving, it will carry on moving at the same velocity. The object could keep a constant velocity, but if the moments on it are not also balanced, it could be made to start rotating. The principle of moments tells us that if the total of all the moments trying to turn an object clockwise is equal to the total of all moments trying to turn an object anticlockwise, then it will be in rotational equilibrium. This means it will either remain stationary, or if it is already rotating it will continue at the same speed in the same direction.
如果我们将作用在一个物体上的所有力相加,并且考虑到它们的方向的合力为零,那么该物体将处于平衡状态。因此,它将保持静止,或者,如果它已经在移动,它将以相同的速度继续移动。物体可以保持恒定的速度,但如果其上的力矩不平衡,则可以使其开始旋转。力矩原理告诉我们,如果试图顺时针转动物体的所有力矩之和等于试图逆时针转动物体的所有力矩之和,那么它将处于旋转平衡状态。这意味着它要么保持静止,要么如果它已经在旋转,它将以相同的速度继续沿相同的方向前进。

A fig D As the metre-long beam is balanced, the sum of all the clockwise moments must equal the sum of all the anticlockwise moments.
A 图 D 由于米长的光束是平衡的,因此所有顺时针力矩的总和必须等于所有逆时针力矩的总和。

LEARNING TIP 学习小贴士

The clockwise moments and the anticlockwise moments must all be taken about the same pivot point.
顺时针力矩和逆时针力矩都必须围绕同一个枢轴点取。

EXAM HINT 考试提示

Note that the steps and layout of the solution in this worked example are suitable for moments questions in the exam.
请注意,此工作示例中解决方案的步骤和布局适用于考试中的 Moments 问题。

WORKED EXAMPLE 工作示例

In fig D, we can work out the weight of the beam if we know all the other weights and distances. The beam is uniform, so its weight will act from its centre. The length of the beam is 100 cm . So if x 1 = 20 cm x 1 = 20 cm x_(1)=20cmx_{1}=20 \mathrm{~cm}, then x 0 x 0 x_(0)x_{0} must be 30 cm , and x 2 = 80 cm x 2 = 80 cm x_(2)=80cmx_{2}=80 \mathrm{~cm}. The dinosaur ( W 1 ) W 1 (W_(1))\left(W_{1}\right) weighs 5.8 N and the toy car’s weight ( W 2 ) W 2 (W_(2))\left(W_{2}\right) is 0.95 N . In equilibrium, principle of moments: sum of anticlockwise moments = sum of clockwise moments
在图 D 中,如果我们知道所有其他重量和距离,我们就可以计算出梁的重量。光束是均匀的,因此它的重量将从其中心作用。横梁的长度为 100 厘米。所以如果 ,则 x 1 = 20 cm x 1 = 20 cm x_(1)=20cmx_{1}=20 \mathrm{~cm} x 0 x 0 x_(0)x_{0} 必须为 30 cm 和 x 2 = 80 cm x 2 = 80 cm x_(2)=80cmx_{2}=80 \mathrm{~cm} 。恐龙 ( W 1 ) W 1 (W_(1))\left(W_{1}\right) 重 5.8 N,玩具车重 ( W 2 ) W 2 (W_(2))\left(W_{2}\right) 0.95 N 。在平衡状态下,力矩原理:逆时针力矩之和 = 顺时针力矩之和
W 1 x 1 = W 0 x 0 + W 2 x 2 5.8 × 0.20 = W 0 × 0.30 + 0.95 × 0.80 W 0 = 1.16 ( 0.76 ) 0.30 W 0 = 1.3 N W 1 x 1 = W 0 x 0 + W 2 x 2 5.8 × 0.20 = W 0 × 0.30 + 0.95 × 0.80 W 0 = 1.16 ( 0.76 ) 0.30 W 0 = 1.3 N {:[W_(1)x_(1)=W_(0)x_(0)+W_(2)x_(2)],[5.8 xx0.20=W_(0)xx0.30+0.95 xx0.80],[:.quadW_(0)=(1.16-(0.76))/(0.30)],[W_(0)=1.3N]:}\begin{aligned} W_{1} x_{1} & =W_{0} x_{0}+W_{2} x_{2} \\ 5.8 \times 0.20 & =W_{0} \times 0.30+0.95 \times 0.80 \\ \therefore \quad W_{0} & =\frac{1.16-(0.76)}{0.30} \\ W_{0} & =1.3 \mathrm{~N} \end{aligned}

CENTRE OF GRAVITY 重心

The weight of an object is caused by the gravitational attraction between the Earth and each particle contained within the object. The sum of all these tiny weight forces appears to act from a single point for any object, and this point is called the centre of gravity. For a symmetrical object, we can calculate the position of its centre of gravity, as it must lie on every line of symmetry. The point of intersection of all lines of symmetry will be the centre of gravity. Fig E illustrates this with two-dimensional shapes, but the idea can be extended into three dimensions. For example, the centre of gravity of a sphere is at the sphere’s centre.
物体的重量是由地球与物体中包含的每个粒子之间的引力引起的。对于任何物体,所有这些微小的重量力的总和似乎都从一个点起作用,这个点称为重心。对于对称物体,我们可以计算其重心的位置,因为它必须位于每条对称线上。所有对称线的交点将是重心。图 E 用二维形状说明了这一点,但这个想法可以扩展到三维。例如,球体的重心位于球体的中心。

together and this answer used to calculate some sort of combined moment.
一起,这个答案用于计算某种组合力矩。

LEARNING TIP You can consider the terms 'centre of gravity' and 'centre of mass' to mean the same thing. They are exactly the same for objects that are small compared
LEARNING TIP You can consider the terms 'centre of gravity' and 'centre of mass' to mean the same thing. They are exactly the same for objects that are small compared
LEARNING TIP You can consider the terms 'centre of gravity' and 'centre of mass' to mean the same thing. They are exactly the same for objects that are small compared
LEARNING TIP You can consider the terms 'centre of gravity' and 'centre of mass' to mean the same thing. They are exactly the same for objects that are small compared
LEARNING TIP You can consider the terms 'centre of gravity' and 'centre of mass' to mean the same thing. They are exactly the same for objects that are small compared
LEARNING TIP You can consider the terms 'centre of gravity' and 'centre of mass' to mean the same thing. They are exactly the same for objects that are small compared to the size of the Earth.
学习提示 您可以将“重心”和“质心”这两个术语视为同一含义。对于比较小的物体,它们完全相同 学习提示 您可以将“重心”和“质心”这两个术语视为同一含义。对于比较小的物体,它们完全相同 学习提示 您可以将“重心”和“质心”这两个术语视为同一含义。对于比较小的物体,它们完全相同 学习提示 您可以将“重心”和“质心”这两个术语视为同一含义。对于比较小的物体,它们完全相同 学习提示 您可以将“重心”和“质心”这两个术语视为同一含义。对于比较小的物体,它们完全相同 学习提示 您可以将“重心”和“质心”这两个术语视为同一含义。对于与地球大小相比较小的物体,它们完全相同。

In order to calculate the sum of the moments in either direction, each individual moment must be calculated first and these individual moments then added together. The weights and/or
为了计算任一方向的力矩之和,必须首先计算每个单独的力矩,然后将这些单独的力矩相加。权重和/或

distances cannot be added
无法添加距离

  • fig E E E\mathbf{E} The centre of gravity of a symmetrical object lies at the intersection of all lines of symmetry.
    E E E\mathbf{E} 图 对称物体的重心位于所有对称线的交点处。

IRREGULAR OBJECTS 不规则对象

The centre of gravity of an irregularly shaped object will still follow the rule that it is the point at which its weight appears to act on the object. A Bunsen burner, for example, has a heavy base. As such, the centre of gravity is low down near that concentration of mass, as there will be a greater attraction by the Earth’s gravity to this large mass.
形状不规则的物体的重心仍将遵循规则,即它是其重量似乎作用在物体上的点。例如,本生灯的底座很重。因此,重心在靠近该质量集中的地方很低,因为地球的引力对这个大质量的吸引力会更大。

  • fig F Balancing a broom on its centre of gravity.
    图 F 在扫帚的重心上保持平衡。


    fig G inding the centre of gravity of an irregular rod (broom).
    图 G 表示不规则杆(扫帚)的重心。

PRACTICAL SKILLS 实践技能

Finding the centre of mass of an irregular rod
找到不规则杆的质心

In this investigation, we use the principle of moments to find the centre of gravity of a broom. It is not easy to estimate/determine the (location of the) centre of gravity by looking at a broom because it is not a symmetrical object. With the extra mass at the brush head end, the centre of gravity will be nearer that end. If you can balance the broom on the edge of a thick metal ruler, then the centre of gravity must lie above the ruler edge. As the perpendicular distance from the line of action to the weight is zero, the moment is zero so the broom sits in equilibrium.
在这项研究中,我们使用力矩原理来找到扫帚的重心。通过观察扫帚来估计/确定重心(位置)并不容易,因为它不是一个对称的物体。由于刷头末端有额外的质量,重心将更接近该末端。如果您可以在粗金属尺的边缘上平衡扫帚,那么重心必须位于尺子边缘上方。由于从作用线到砝码的垂直距离为零,因此力矩为零,因此扫帚处于平衡状态。

You will probably find it difficult to balance the broom exactly, so you can use an alternative method. First you measure the mass of the broom ( M ) ( M ) (M)(M) using a digital balance. Then you use a set of hanging masses (of mass m m mm ) to balance the broom more in the middle of the handle, as in fig G . When the broom is balanced, you measure the distance ( d ) ( d ) (d)(d) from the hanging masses to the pivot. You calculate the distance ( x ) ( x ) (x)(x) from the pivot to the centre of gravity of the broom using the principle of moments:
您可能会发现很难精确平衡扫帚,因此您可以使用其他方法。首先,使用数字天平测量扫帚 ( M ) ( M ) (M)(M) 的质量。然后你使用一组悬挂质量(质量 m m mm )在手柄中间更平衡扫帚,如图 G 所示。当扫帚平衡时,您可以测量从悬挂质量到枢轴的距离 ( d ) ( d ) (d)(d) 。您可以使用弯矩原理计算从枢轴到扫帚重心的距离 ( x ) ( x ) (x)(x)

clockwise moment = anticlockwise moment
顺时针力矩 = 逆时针力矩
m g × d = M g × x x = m d M m g × d = M g × x x = m d M {:[mg xx d=Mg xx x],[x=(md)/(M)]:}\begin{aligned} m g \times d & =M g \times x \\ x & =\frac{m d}{M} \end{aligned}
Note: Do not get into the habit of using only the mass in moments calculations, as the definition is force times distance. It just happens that in this case g g g\boldsymbol{g} cancels on each side.
注意:不要养成只使用力矩质量计算的习惯,因为定义是力乘以距离。在这种情况下,恰好每一侧都 g g g\boldsymbol{g} 取消。
Safety Note: Securely clamp the base of the stand to the bench. Do not use a very heavy broom as you would need a heavier counterweight which could fall and cause injury.
安全说明:将支架底座牢固地夹在工作台上。不要使用非常重的扫帚,因为您需要更重的配重,这可能会掉落并造成伤害。

CHECKPOINT 检查站

  1. What is the moment of a 252 N force acting on a solid object at a perpendicular distance of 1.74 m from an axis of rotation of the object?
    在距物体旋转轴垂直 1.74 m 处作用在固体物体上的 252 N 力的力矩是多少?
  2. A child and his father are playing on a seesaw, see fig H. They are exactly balanced when the boy (mass 46 kg ) sits at the end of the seesaw, 2.75 m from the pivot. If his father weighs 824 N , how far is he from the pivot?
    一个孩子和他的父亲在跷跷板上玩耍,见图 H。当男孩(质量 46 公斤)坐在跷跷板的末端,距离枢轴 2.75 m 时,它们完全平衡。如果他父亲的体重是 824 N,他离枢轴有多远?


    fig H 图 H
  3. The weight of the exercise book in the left-hand picture in fig B B BB causes a rotation so it moves towards the second position. Explain why it does not continue rotating but comes to rest in the position of the second picture.
    图中左侧图片中练习簿的重量 B B BB 导致旋转,因此它向第二个位置移动。解释为什么它不继续旋转,而是停在第二张图片的位置。
  4. If the same set-up as shown in fig D was used again, but the toy car was replaced with a banana weighing 1.4 N , find out where the banana would have to be positioned for the beam to balance calculate the new x 3 x 3 x_(3)x_{3}.
    如果再次使用图 D 所示的相同设置,但玩具车被重 1.4 N 的香蕉取代,找出香蕉必须放置在何处才能使横梁平衡,计算新的 x 3 x 3 x_(3)x_{3}

SUBJECT VOCABULARY 学科词汇

equilibrium the situation for a body where there is zero resultant force and zero resultant moment. It will have zero acceleration
Equilibrium 合力为零和合力矩为零的物体的情况。它的加速度为零

principle of moments a body will be in equilibrium if the sum of clockwise moments acting on it is equal to the sum of the anticlockwise moments
力矩原理 如果作用在物体上的顺时针力矩之和等于逆时针力矩之和,则物体将处于平衡状态

centre of gravity the point through which the weight of an object appears to act
重心 物体的重量似乎通过其作用的点