Does This Material Follow Ohm's Law?

AI Thread Summary
The discussion revolves around whether a material follows Ohm's Law based on its current and voltage responses, with initial observations suggesting it does not obey the law due to varying current with increased voltage. The user also seeks clarification on calculating power consumption for a light bulb and the time required to use one kilowatt-hour of energy. While the user correctly calculates the power as 60W, there is confusion regarding the energy conversion and time calculations. Participants encourage the user to work through the problems independently, emphasizing the importance of understanding the underlying concepts. The thread highlights the collaborative nature of learning in physics.
physicsphobic
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urgently needed help on intro physics...

This is my first time posting on here so I hope someone will answer my questions.

Here they are:

1. A scientist testing a material discovers that when a voltage of 10V is placed across a material a current of 10A flows. When a voltage of 20V is placed across it, however, a current of 15A flows through it. Does this material obey Ohm's law? Give reason for your answer.

2.a. How much power does a light bulb connected to a 120V outlet use if if draws 0.5A of current? My answer is: 120V *0.5A= 60W
b. One kilowatt-hour is a measure of energy which is equal to 3,600,000J. How many hours would you have to leave on the light bulb from part a in order to expend one kilowatt-hour of energy?

c. If one kilowatt-hour costs $.10, how long would you have to leave the light bulb from part a on in order to spend $1.00 on electricity?

3. The electric charge of an electron is -1.6x10^-19C. What is the force exerted between two electrons separated by one meter? (Be sure to note whether the force is attractive or repulsive).

Thanks for any help given to this newbie!

Joanne
 
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Welcome to PF!

Hi Joanne! Welcome to PF! :wink:

Your answer to 2a is correct.

For the others, show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 


OK, since nobody wants to give me the answers I'll try to give what I think are the answers and, hopefully someone will be kind enough to give me the correct answers.

Since 2a is correct, here's what I think about 2b:
according to my textbook I use this formula..energy=power * time.
plugging the 3,600,000J in for energy and 60W in for power I get: 3600000=60W * time
leaving time on one side, time=60W/3600000=60,000 hours? Is this right because my algebras a little rusty.

2c has me stumped so any help is greatly appreciated.
The first question I think it does obey ohms law because the greater the resistance, the less the current and if the resistance is doubled the current will be half what it would be otherwise. Right??
 
Hi Joanne! :smile:

Sorry, but on this forum we don't give you the answers. :wink:
physicsphobic said:
Since 2a is correct, here's what I think about 2b:
according to my textbook I use this formula..energy=power * time.
plugging the 3,600,000J in for energy and 60W in for power I get: 3600000=60W * time
leaving time on one side, time=60W/3600000=60,000 hours? Is this right because my algebras a little rusty.

Why are you using 3,600,000?

The question asks for 1 kWhr, (that's the same power as a kW heater for an hour) … joules have nothing to do with it.

Try again. :smile:
The first question I think it does obey ohms law because the greater the resistance, the less the current and if the resistance is doubled the current will be half what it would be otherwise. Right??

Ohm's law mentions the amount of the resistance … how much is the resistance in this case? :smile:
 


tiny-tim said:
Sorry, but on this forum we don't give you the answers. :wink:

We also have forums specifically for getting help with homework / coursework exercises. They're sub-forums of "Homework & Coursework Questions" which is near the top of the list of forums. I've moved this thread to "Introductory Physics."
 


figured out most of the problems myself so thanks anyway, tiny-tim for your "help".
 
physicsphobic said:
figured out most of the problems myself so thanks anyway, tiny-tim for your "help".

you're welcome :wink: … we always think our "help" is most successful when the OP ends up answering the question him/herself :biggrin:
 
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