Recent content by ToTalk

Thanks for the corrections. In the question it says the wire starts at 20 C, I just forgot to include that on here. I find the melting point of copper to be 1084.62 C, so this wire would be mush! Thanks so much for your help, I really learned a lot, and I really appreciate it.

Thanks again for your help. So if I use R = (1.68E-8) * (1m / 7.85E-7 m^2) = .021 ohms / meter Then I take P = I^2R = 400^2 * .021 = 3360 Watts / Meter? Energy = 3360 Joules / second / meter Copper Specific Heat = 390 J / kg *C Volume of 1 meter = 7.85E-7 * 1 = 7.85E-7 cubic...

Thank you for your help! :) Ok, I understand what you are saying, but I am not sure how to get power from what I am given? Do I multiply the resistivity by the area 1.68E-8 x 7.85E-7 = 1.319E-14 ohms / meter? 1.319E-14 ohms / meter x 400 A = 5.278E-12?

1. Homework Statement - Consider a copper wire 1 mm in diameter providing the power to run an appliance drawing 4.8 kW at 12 V. Assuming no heat radiated away from the wire while the currecnt flows, (a) what will the temperature of the wire be after the current has run for 1 second through...

Ok I got it. Thanks so much for all your help tonight OlderDan. I really appreciate it. I better run to bed so that I can be at least a little bit awake for tomorrow!

Ok so when the positive charged insulator is near the closer sphere, it will pull the electrons from both spheres toward the closer sphere and push the protons from both spheres toward the further one, and then when the spheres are separated, the sphere on the right will maintain a positive due...

Oh ok. I just did that in excel and it worked. Thanks again, that was interesting and very educational. I really gained some understanding from your help.

Hi, Any help on this would be appreciated. 1. If you bring a positively charged insulator near one of two metallic spheres that are in contact and then separate the spheres, what charge will the sphere further away from the insulator have? 2. Homework Equations - none 3. I...

Ok, so as t changes 1, x must change 25 / 1.9 = 13.16? So the answer is x changes 13.16 m for every 1 s change in t. Is that correct? It seems right based on what you explained. Thanks again for all your help OlderDan.

If anyone else knows how to help with this, I would appreciate the help. I think that when 1.9x - 25t = pi/2, the maximum amplitude is reached, but I am not sure how to use this info. Thanks

Hi OlderDan, Thanks for the help! Ok, I realize that what is inside the parentheses must equal pi / 2. And does that mean that the maximum y value is 2.3? I'm still not sure about the next part. Thanks

1. Consider a transverse wave represented by the function y = 2.3 sin(1.9x - 25t) where y, x, t are in meters, meters, and seconds Determine the speed in m/s 2. No relevant equations. 3. I don't understand the way this problem is written. It seems like...

Just got it. For some reason I distributed d to both terms in the numerator instead of f_{2} . Thanks for your help.

Hi Kurdt, Thanks for the info. Are you talking about this: \frac{f_2 (d - f_1) } { d - (f_1 +f_2) } If so, I had seen that, but I am unsure of how to get to that from the given equation. Thanks

Hi, I have been working on this for a little while and think I am approaching it incorrectly. If someone can help, I'd really appreciate it. This is for a basic introductory calculus based Physics class that I am in. 1. Mathematically demonstrate the following equation with no numerical...