Recent content by cwesto

Homework Statement A capacitor is discharged through a 100\Omega resistor. The discharge current decreases to 27.0% of its initial value in 2.80ms. What is the value of the capacitor? Homework Equations Qf=Qi*e(\frac{-t}{CR}) The Attempt at a Solution ln(.27)=\frac{.0028s}{C100}...

I=V/R Rtotal=12+18 I=12/30 PR1=RI2 PR1=12*\frac{2}{5}*\frac{2}{5} PR1=48/25 PR2=\frac{72}{25}=2.88 Thanks!

Homework Statement Homework Equations P=IV I=V/R P=(V/R)V R=V2/R The Attempt at a Solution PR1: 122/12=12 PR2: 144/18 I'm not sure what to do from here. Thanks for any input and help!

approx. 30.44 days in a month 30.44*24=730.56 hours in a month. 800000Wh/730.56h = 1095 W! oh... lol. DUH! Thanks! :D

so how do I get from kWh to W?

Homework Statement A typical American family uses 800 kWh of electricity a month. What is the average current in the 120 V power line to the house? Homework Equations P=IV The Attempt at a Solution 800kW=800000W 800000=120I I=\frac{800000}{120} I'm getting this answer as...

wait sorry they don't give the current my bad. i was looking at another problem. sorry all the information up there is what they give me to solve the problem, nothing more.

the current is 2 amps. what is the equation to find the voltage drop?

Homework Statement How much power is dissipated by the 12\Omega resistor in the figure? PR1=______W How much power is dissipated by the 18 resistor in the figure? PR2=______W Homework Equations P=IV I=\frac{V}{R} The Attempt at a Solution P=\frac{V}{R}*V P=V2/R...

Oh sorry.

The problem: f(x,y)=-\frac{-7x-2y}{9x+7y} find: fx(x,y) fy(x,y) The attempt: fx(x,y)=\frac{-7-2y}{9+7y} fy(x,y)=\frac{-7x-2}{9x+7} Questions: I'm not exactly sure how to find the partail derivative with a fraction like this one.

Oh yeah! ΔV = ½mv²/q Your right. Thanks.

I'll ask my professor then post. I have class now. Thanks though.

Wouldn't "ΔV = ½mv²/q" give me ΔV while the speed is 4.70×105m/s? I want to know the ΔV when the speed is 0.

The distance in which they want the electron to stop.