物理论文英文版

物理论文英文版

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平抛运动

物体以一定的初速度沿水平方向抛出，如果物体仅受重力作用，这样的运动叫做平抛运动。平抛运动可看作水平方向的匀速直线运动以及竖直方向的自由落体运动的合运动。平抛运动的物体,由于所受的合外力为恒力,所以平抛运动是匀变速曲线运动,平抛物体的运动轨迹为一抛物线。

平抛运动的时间仅与抛出点的竖直高度有关；物体落地的水平位移与时间（竖直高度）及水平初速度有关。

平抛运动可用两种途径进行解答 . 一种是位移途径; 另一种是速度途径.

位移途径为:

L(水平)=vt L(竖直）= 1/2gt^2

还有速度途径为：

t=v/t v（竖直）=gt

即可求解

平抛运动的分析

平抛运动实际上是以下两个运动的合运动：

（1）在水平方向上不受外力，所以做匀速直线运动，其速度为平抛运动的初速度；

（2）在竖直方向上，物体只受重力作用，所以做自由落体运动。

这两个分运动各自独立，又是同时进行，具有分运动的独立性和等时性。

(3) 平抛运动的运动轨迹: ∵x=v0t,H=1/2gt^2

∴ X2=H（2V0∧2）／g 为二次方程

∴其运动轨迹为抛物线。

平抛运动的规律

公式:水平方向:s=v0*t

竖直方向:h=1/2gt^2

两个公式中时间t是相同的

和速度公式√{V0^2+(gt)^2}

1．运动时间只由高度决定

设想在高度H处以水平速度vo将物体抛出，若不计空气阻力，则物体在竖直方向的运动是自由落体，由公式可得： h=1\2gt^2，由此式可以看出，物体的运动时间只与平抛运动开始时的高度有关。t=√(2h/g)

2．水平位移由高度和初速度决定

平抛物体水平方向的运动是匀速直线运动，其水平位移，将代入得：S(水平)=v0t=v0√(2h/g)

，由此是可以看出，水平位移是由初速度和平抛开始时的高度决定的。

3．在任意相等的时间里，速度的变化量相等

由于平抛物体的运动是水平方向的匀速直线运动和竖直方向的自由落体运动的和合运动。运动中，其水平运动的速度保持不变，时间里，水平方向的分速度的变化量为零，竖直方向的分速度的变化量为9.8m／s^2，而时间里合速度的变化量为两个方向速度变化量的矢量和，其大小为：，方向竖直向下。由此可知，在相等的时间里，速度的变化量相等，由此也可以知道，在任意相等的时间里，动量的变化量相等。

4．任意时刻，速度偏向角的正切等于位移偏向角正切的两倍

5．任意时刻，速度矢量的反向延长线必过水平位移的中点

6．从斜面上沿水平方向抛出物体，若物体落在斜面上，物体与斜面接触时的速度方向与水平方向的夹角的正切是斜面倾角正切的二倍

7．从斜面上水平抛出的物体，若物体落在斜面上，物体与斜面接触时速度方向与斜面的夹角与物体抛出时的初速度无关物体落在斜面上时，速度方向与斜面的夹角与初速度无关，只取决于斜面的倾角。

钟摆运动

简单来说，当不受外力干扰时，钟摆所呈现的状态是静止不动的，一旦受到外力影响，钟摆就会开始左右晃动，受到的外力愈大，受惯性的影响，摆回来的力道也会愈强，随著地心引力的影响而愈摆愈慢，最后终於回到原点呈现原本静止的状态。

概述

加速度（Acceleration）是速度变化量与发生这一变化所用时间的比值。是描述物体速度改变快慢的物理量，通常用a表示，单位是m/s^2（米每二次方秒）。加速度是矢量，它的方向是物体速度变化量的方向，它的方向与合外力的方向相同，其方向表示速度改变的方向，其大小表示速度改变的大小。地球上各个地方的加速度都是不同的。牛顿运动学第二定律认为，a=F/m， F为物体所受合外力，m为物体的质量。力是改变物体运动状态的条件，而加速度则是描述物体运动状态的物理量。加速度与速度无必然联系，加速度很大时，速度可以很小，速度很大时，加速度也可以很小。从微分的角度来看，加速度是速度对时间求导，是v－t图像中的斜率。当加速度与速度方向在同一直线上时，物体做变速直线运动，如汽车以恒定加速度启动[1]（匀加速直线运动），简谐振动（变加速直线运动）；当加速度与速度方向不在同一直线上时，物体做变速曲线运动，如平抛运动（匀加速曲线运动），匀速圆周运动（变加速曲线运动）；加速度为零时，物体静止或做匀速直线运动。任何复杂的运动都可以看作是无数的匀速直线运动和匀加速运动的合成。我们还应用极限的思想去思考加速度的问题。

公式

s=v0t+1/2a(t^2) ( v0是初速度，t是时间)

a=(V-V0)/(t-t0)=△ V/△ t

V=Vo+aX△t

2a△x=v2-v02加速度的实例1、匀加速运动。

曲线运动

曲线运动中的加速度计算公式：

a=rω^2=v^2/r

和圆周运动一样 ， 这种加速度称为向心加速度是物体的合外力，也指向圆心。

但是加速度保持不变的时候，物体也有可能做曲线运动。

比如，当你把一个物体沿着水平桌面往前使劲一推（即物体离开桌面时做平抛运动），你会发现，这个物体离开桌面以后，在空中划过一条曲线，落在了地上。

物体在你松开手以后，且离开桌面后，受到的只有重力。重力永远是竖直向下的，因此加速度的方向也是竖直向下的，且大小也不改变。但是物体离开桌面的时候，仍然具有惯性，因此想保持继续平行前进。这个时候，物体的速度方向与加速度方向就不在同一直线上了。物体就会往力的方向偏转，划过一条往地面方向偏转的曲线。

但是这个时候，由于重力大小不变，因此加

变加速运动

我们首先一定要清楚。加速度是一个矢量。它有大小与方向。比如，一个人从背后以1N的力推你，和一个人站在你正前方，以同样大小的力推你，你倒下去的方向是不一样的。

因此，假如在一个运动中，加速度的大小保持不变，但是方向在变化，这就不再是匀加速运动，而是变加速运动了。比如匀速圆周运动。

匀速运动  速度（速率、方向）不变的运动

也可以说是加速度为0的运动

匀速运动一定是匀速直线运动 .

速度是矢量，包括大小和方向，匀速运动是速度不变的运动，运动方向不发生改变，所以匀速运动一定是匀速直线运动！

定义：单位时间内通过相同的路程的运动叫做匀速运动。

2．匀速直线运动，速度、速率、位移公式S=υt，S～t图线，υ～t图线

3．变速直线运动，平均速度，瞬时速度

匀变速直线运动的基本规律及重要推论

（1）匀变速直线运动的基本规律通常是指所谓的位移公式和速度公式

S=υ0t+ at2

υ=υ0+at

（2）在匀变速直线运动的基本规律中，通常以初速度υ0的方向为参考正方向，即υ0＞0此时加速度的方向将反映出匀速直线运动的不同类型：

①a＞0，指的是匀加速直线运动；

②若a=0，指的是匀速直线运动；

③若a=0，指的是匀减速直线运动。

（3）匀变速直线运动的基本规律在具体运用时，常可变换成如下推论形式

推论1： υ2－ =2as

推论2： = （υ0+υ）

推论3：△S=a△T2

推论4： = (υ0+υ)

推论5： =

推论6：当υ0=υ时，有

S 1：S2 ：S3：……=12 ：22 ：32 ：……

SⅠ ：SⅡ ：SⅢ ：……=1 ：3 ：5 ：……

υ1 ：υ2 ：υ3：……=1 ：2 ：3 ：……

t1 ：t2 ：t3 ：……=1 ：( －1) ：( － ) ：……

3．匀变速直线运动的υ－t图

用图像表达物理规律，具有形象，直观的特点。对于

匀变速直线运动来说，其速度随时间变化的υ～t图线如图

1所示，对于该图线，应把握的有如下三个要点。

（1）纵轴上的截距其物理意义是运动物体的初速度υ0；

（2）图线的斜率其物理意义是运动物体的加速度a；

（3）图线下的“面积”其物理意义是运动物体在相应

的时间内所发生的位移s。                              图1

4．竖直上抛运动的规律与特征。

（1）竖直上抛运动的条件：有一个竖直向上的初速度υ0；运动过程中只受重力作用，加速度为竖直向下的重力加速度g。

（2）竖直上抛运动的规律：竖直上抛运动是加速度恒定的匀变速直线运动，若以抛出点为坐标原点，竖直向上为坐标轴正方向建立坐标系，其位移公与速度公式分别为

S=υ0t－ gt2

υ=υ0－gt

（3）竖直上抛运动的特征：竖直上抛运动可分为“上升阶段”和“下落阶段”。前一阶段是匀减速直线运动，后一阶段则是初速度为零的匀加速直线运动（自由落体运动），具备的特征主要有：

①时间对称——“上升阶段”和“下落阶段”通过同一段大小相等，方向相反的位移所经历的时间相等，即

t上=t下

②速率对称——“上升阶段”和“下落阶段”通过同一位置时的速率大小相等，即

υ上=υ下

不好意思 差不多 只能这样了 你再归纳一下

图传不上来

需要写一篇物理的小论文（科目：ENERGY AND THE ENVIRONMENT ），要求如下 用中英文回答皆可....感激不尽

An Introduction To Electrical Power And Energy
Why Do You Need To Know About Electrical Energy?
What Is Electrical Power?
Power and Energy In Electrical Devices
Resistors
Batteries
Problems

You are at: Basic Concepts - Quantities - Power & Energy
Return to Table of Contents

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Why Do You Need To Know About Electrical Energy?
A power station is a place where other forms of energy - coal, gas, potential energy in water and nuclear energy - are turned into electrical energy for transmission to places that use electrical energy. Electrical engineering is concerned with transmission and ultilization of two things - energy and information. Here, in this lesson, we are going to focus on power and energy. In this lesson you will want to learn the following.

Given an electrical circuit or device
Be able to compute instantaneous rate of energy use (power).
Be able to compute how much energy is used over a period of time.
Be able to compute how much energy is stored in an electrical storage device like a battery or a capacitor.

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What Is Electrical Power?

Electrical power is conceptually simple. Consider a device that has a voltage across it and a current flowing through it. That situation is shown in the diagram at the right.
The voltage across the device is a measure of the energy - in joules - that a unit charge - one couloumb - will dissipate when it flows through the device. (Click here to go to the lesson on voltage if you want to review.) If the device is a resistor, then the energy will appear as heat energy in the resistor. If the device is a battery, then the energy will be stored in the battery.
The current is the number of couloumbs that flows through the device in one second.j (Click here to go to the lesson on current if you want to review.)
If each couloumb dissipates V joules, and I couloumbs flows in one second, then the rate of energy dissipation is the product, VI.
That's what power is - the rate at which energy is expended. The rest of the story includes these points.
It doesn't matter what the electrical device is, the rate at which energy is delivered to the device is VI as long as the voltage and current are defined as shown.
The power can be negative. If the device is a battery, then current - as defined in the figure - can easily be negative if, for example, a resistor is attached to the battery. If the power is negative, then the rate at which the device expends energy is negative. That really means that it is delivering energy in that situation.

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Power in Electrical Devices
A resistor is one device for which you can compute power dissipation.

A symbol for a resistor is shown below, along with a voltage, Vr, across the resistor and a current, Ir, flowing through the resistor.

We can compute the power delivered to the resistor. It's just the product of the voltage across the resistor and the current through the resistor, VrIr.
But there's more to the story.
In a resistor, there is a relationship between the voltage and the current, and we can use that knowledge to get a different expression - one that will give more insight.
We know that Vr = Rir, so the power is just:
Power into the resistor = VrIr = (RIr)Ir = R(Ir)2.
We can also use the expression for the current Ir = Vr/R,
Power into the resistor = VrIr = Vr(Vr/R) = (Vr)2/R.
At different times, these two results - which are equivalent - can be used - whichever is appropriate. Besides being a useful result tthese are also illuminating results (And that's not a reference to the fact that a typical light bulb is a resistor that dissipates power/energy.).
The power dissipated by a resistor is always positive. That means that it does not (and in fact it could not) generate energy. It always dissipates energy - uses it up - contributing to the heat death of the universe.
We know the power is positive because R is always positive (and it will always be for any resistor that doesn't have hidden transistors) and because the square of the current has to be a positive number.

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Problems
P1. You have a 1KW resistor, and there is 25 volts across the resistor. Determine the power (in watts) that the resistor dissipates.

Enter your answer in the box below, then click the button to submit your answer. You will get a grade on a 0 (completely wrong) to 100 (perfectly accurate answer) scale.
You have a 1KW resistor, and there is 25 volts across the resistor. Determine the power (in watts) that the resistor dissipates.

Your grade is:

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P2. You have a 25 watt light bulb that operates with 12.6 volts across it. Determine the resistance of the light bulb.
Enter your answer in the box below, then click the button to submit your answer.

Your grade is:

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Power In Batteries
Batteries are ubiquitous components. They are in TV remotes, cell phones and things like that. But, batteries also appear in places you don't expect them to be. For example, you can turn this computer off. When you turn it back on it remembers things and recalculates things like the time. Now, you expect that for things that can be stored on a hard drive. You don't expect it for the time.

When you turn this computer off and later turn it back on it will have the right date and time. How does it do that? If you think about it (and don't do that for too long!) you have to believe that there is a battery somewhere inside the computer and that when you turn the computer off that battery runs some sort of little clock hidden inside the computer. You can't see the clock and you wouldn't even know it's there, but you can probably see the time now on the task bar of this computer - and it's probably close to being right!

Batteries are used to solve many problems.

They are used to provide power to run things like computer clocks that need to keep running even in the absence of AC power.
They are used to store energy for things like starting a car. When you run the car you generate energy (from the gasoline) and store it in the car battery. Then there is energy there when you need it to get the car going again.
They are used for low power devices to make them portable. That includes things like cell phones, TV remotes and calculators.
You use batteries - whether you want to or not, and whether you know it or not! You need to be able to compute some of the quantities involved. Here is a simple circuit where a resistor is connected to a battery. We know some salient facts about this circuit.
There is energy stored in the battery and the battery delivers stored energy to the resistor.
The resistor dissipates energy, and what happens physically is that the electrical energy that is delivered to the resistor gets turned into heat energy and the resistor becomes warmer.
Now, we need to look at a circuit diagram for this situation. That circuit diagram is shown to the right of the picture below.

In the diagram, we have defined a battery voltage, Vb, and a current, Ir.
Notice that we have used a natural definition for the current polarity. We have the arrow pointing out of the battery and into the resistor. We do that because we know that positive charge actually flows from the battery terminal through the resistor.
That definition of current polarity raises questions about calculation of power to/from the battery. Let us consider the power flow into the battery.

Power flow into the battery or any other device - is the product, VI, when
V is the voltage across the device, and
I is the current flowing into the device.
Remember, for our polarity conventions here, the current arrow points into the terminal of the device that is labelled "+" for the voltage definition.
We have reproduced the diagram from above here to emphasize how the voltage and current polarities are defined. Notice that the current arrow in our earlier definition points toward the "+" sign on the device.

In the battery-resistor circuit below, the current arrow is directed out of the positive ("+") terminal of the battery. That means the power delivered to the battery must be computed by (note the minus sign!):

P = - VI

What does this mean? Let's look at a numerical example. Let's assume the battery voltage is 12 volts and the resistor is 24 ohms. That means the current is 0.5 amps, i.e.:

Ir = 12v/24W = 0.5a

In other words, the power flowing into the battery is:

P = - Vb Ir = - 12 * 0.5 = - 6w

The power flowing into the battery is negative!
The power flowing out of the battery is positive!
And, it makes sense because we know the battery supplies power.
How much energy is stored in a battery?
Batteries are often rated in ampere-hours (or milliampere-hours) and an ampere-hour is really a unit of charge.
As a battery is used it discharges - charge flows from the battery - but it tends to hold a constant voltage. This is different than the internal resistance of the battery. What we are saying here is that as time goes on - for the same current drawn from the battery - the voltage stays about the same. There may be a slight drop-off but it is not very large.
Thus, if we have a 12.6 v battery, it will have something close to 12.6v until it gets close to being discharged.
Let's say we have a 12.6v battery rated at 70 ampere-hours.
Assuming it can deliver 1 ampere for 70 hours, then it will be delivering
Power = 12.6v x 1.0 amp = 12.6 watts for 70 hours.
That works out to 12.6 x 70 = 882 watt hours or .882 kw-hr. - and remember you pay the
electric company by the kilowatt-hour!
In joules we have 882 w-hr x 3600sec/hr = 3,175,200 joules.
That might sound like a lot, but an interested student might want to compare that amount of energy with the energy stored in a gallon of gasoline.

物理老师让我们写1000字英文论文.我写柏拉图,他和物理有什么关系呢?

⊙﹏⊙b汗，一千字的
英文
物理
论文
，你老师是外国人啊！
再说了，写
柏拉图
不靠谱。虽然
人们
一直形容中国的
哲学
是学文的，
古希腊
的是学理的。但作为“三哲”之一，柏拉图还真是没给物理一点贡献。最大的贡献，就是他的
数学
哲学观
，说是为数学还过得去，为物理还真没有。他的师父
苏格拉底
也一样，
弟子
亚里士多德
终于有了些发现，但很多都是错的，但是勇于探索的
精神
还是值得赞扬的。在这里可以发现，古希腊学理的过程，先是数学的哲学观，再是联系生活，发现一些生活中的数学关系（物理的雏形，要知道那时
物理学
是包括在统一的
自然哲学
之中的，我国的
墨子
和希腊的亚里士多德算最早的一批人了），然后就是
阿基米德
了，他在物理学上的贡献就不说了，很多。总结下来说，物理的真正
起源
是在柏拉图后。如果真要和柏拉图拉上关系，那就是他架设了研究物理的基础，数学和哲学（物理是充满哲学的，所以很多人都说：“哲学是思想之源。”）！
现在我才发现，这问题是属于哲学的！！！！！！！！平时来这都不来的，大概
你的问题
中有了“物理”两个字，我才随便乱点点进来的吧！下次这样的问题问物理那边去，基本学物理的都知道这样一段历史.（物理书上不要太详细哦!)
⊙﹏⊙b汗啊！