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爱欧尼亚荒原
首页 > 期刊论文 > 物理论文英语题目

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Me馍馍27

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大概就是说 Cu(II)与方形三角骨 (N2S)CuCl2 结合 或者发生反应Jahn–Teller 就会发生形变 还会对光谱和结构的性能 产生随后的影响 chromophore 是化学里的一种 发色团 姜-泰勒效应(Jahn-Teller effect)大概就是要你 研究分析一下 姜-泰勒效应(Jahn-Teller effect), 注意一下:姜-泰勒效应并不能预测变形的方向,只能预测存在一个不稳定的构型。 在试验上,姜-泰勒效应可以通过无机化合物的紫外-可见光谱来研究和解释。所以 论文题目可能是要你用Cu(II)和(N2S)CuCl2 来分析姜-泰勒效应(Jahn-Teller effect), 不知理解对不对,希望对你有帮助!

221 评论

力力力力力

铜(II)配合物N2S金字塔(广场)CuCl2 Jahn-Teller团:阐述了光谱畸变和随后的作用、结构性能

237 评论

木雨大大雯

需要发表吗 我们可以代发的

207 评论

甲壳小咪

你是在美国读书吧。。。看着题目就像。。。typed and double space什么的。。。你最好还是自己做拉。。。第三部分要你读section2的。。。要是你真的不想写给我分我可以帮忙咯。。。我在美国读完高中现在在大学。。。要是你采用其他人的我保证你得C-或者不合格拉。。。那些机器翻译的你还不如自己写...anyways if ur interested in letting me help u just shoot me a message online...good luck with ur essay thou...

151 评论

反叛的路鲁修

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 ProblemsYou are at: Basic Concepts - Quantities - Power & Energy Return to Table of Contents --------------------------------------------------------------------------------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 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 (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. --------------------------------------------------------------------------------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. --------------------------------------------------------------------------------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:--------------------------------------------------------------------------------P2. You have a 25 watt light bulb that operates with 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:--------------------------------------------------------------------------------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 = - VIWhat 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 amps, .: Ir = 12v/24W = other words, the power flowing into the battery is: P = - Vb Ir = - 12 * = - 6wThe 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 v battery, it will have something close to until it gets close to being discharged. Let's say we have a battery rated at 70 ampere-hours. Assuming it can deliver 1 ampere for 70 hours, then it will be delivering Power = x amp = watts for 70 hours. That works out to 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.

350 评论

设计监理

你要写一份2 - 3页(双倍行距,键入的)论文题目:“可以在多大程度上水电预期迎接我们电能需要 未来几十年里吗?” 这篇论文应该有3个部分: 1。介绍了专业术语的定义明确的问题,包括了简要说明物理过程参与蓄能作为重力位能在水中大坝后面并转到电能。 2。一个中间一段提出有关的事实问题

244 评论

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