Recent content by flower76

Sorry I'm confused, you said "Other way around" but based on what you wrote I have it correct that the voltage decreases by k, ie V = Vo/k

Am I overthinking this problem? A 15uF capacitor is connected ot a 55V battery and becomes fully charged. The battery is removed and the circuit is left open. A slab of dielectric material is inserted to completely fill the space between the plates. It has a dielectric constant of 4.8...

Alrighty based on the equation: \frac{kQq}{0.55 - x} = E_{k} + \frac{kQq}{0.55} I found that x is equal to 0.32m, which is the distance that the particle moves before stopping. Does this look a little better? This number seems more appropriate to me.

So I am really confused now. The equation should read: \frac{kQq}{0.55} - \frac{kQq}{x} = E_{k}? And Ek is \frac{kQq}{0.55} + 1/2mv^2? Therefore \frac{kQq}{0.55} - \frac{kQq}{x} = \frac{kQq}{0.55} + 1/2mv^2 ?

So I plugged in the numbers based on my final equation and got 1.05m. So this means that the fired particle goes past the fixed charge before it would turn back, does this make sense?

I'm not sure, so if \frac{kQq}{0.55 - x} = E_{k} then Ek is the PE + KE of q? So my equation would be: \frac{kQq}{0.55 - x} = \frac{kQq}{0.55} + 1/2mv^2?

r is the initial distance between the charges = 0.55m

So here's what I've come up with: Q = fixed charge q = moving charge kQq/(x-0.55m) = kQq/r +(1/2mv^2) Solving for x. I'm not sure if x-0.55 is correct? Or even if I've set up the equation correctly?

Another problem I'm struggling with this one I don't have a clue where to start. Any hints to a starting point would be greatly appreciated. A charge of -4.00uC is fixed in place. From a horizontal distance of 55.0cm a particle of mass 2.50 x 10^-3 kg and charge -3.00uC is fires with an...

Ok I think I got it I added the potenial energy between each of the 6 charge sets and got 4.9x10^7J, hopefully that sounds about right. I think I was getting things confused with the second part of the question: Choose one way of assembling the charges and calculate the potential at each...

I'm not sure that I really understand this question: How much energy is needed to place four positive charges, each of magnitude +5.0mC, at the vertices of a square of side 2.5cm? what I was thinking is that V = kQ/r And since all the charges are equal, and the same distance apart...

If e is 0.312 then 0.312 = (Qh-Ql)/Qh 0.312 = (60 kJ/s - Ql)/60 kj/s Ql = 41.3 kj/s So it exhausts at a rate of 41.3 kj/s Does this look right?

I'm not sure if I'm interpreting this question correctly A heat engine operated between 40C and 380C. Being a real engine, it efficiency is only 60% of that theoretically possible for a Carnot engine at these temperatures. If it absorbs heat at a rate of 60kW at what rate does it exhaust...

I have to hand my assignment in tomorrow and would really like to get this question, can anyone tell me if it is the mean free path that I'm supposed to be finding and if it looks right? I then took the number for mean free path and divided by the velocity I previously found and got 1.06 x...

so I've been playing around with this questionand I think that I need to figure out average distance between molecules. So far I have only determined the volume/molecule and the radius of them based on this. So if I use the mean free path equation I think I should have all the variables to...