Recent content by runfast220

Its a converging lens so does that make the di a virtual image thus making it -9.0m?

ok so... 1/f = 1/infin + 1/6 so f=6.0 1/di = 1/6 - 1/18 = .11 di = 9.0 m hi/ho = - di/do hi/2 = -9/18 hi = -1.0 m does that look right?

Homework Statement The far point of a nearsighted person is 6.0 m from her eyes, and she wears contacts that enable her to see distant objects clearly. A tree is 18.0 m away and 2.0 m high. (a) when she looks through the contacts at the tree, what is its image distance? (b) How high is the...

Homework Statement The drawing shows a crystalline quartz slab with a rectangular cross-section. A ray of light strikes the slab at an incident angle of 1=34o, enters the quartz and travels to point Po (Figure2). This slab is surrounded by a fluid with a refractive index n...

Homework Statement The drawing shows a charged particle (q=2.80x10^-6C) moving along the +y axis with a speed of 4.80X10^6 m/s. A magnetic field of magnitude 3.35x10^-5 T is directed along the +z axis, and an electric field magnitude 123 N/C points along the -x axis. Determine the (a)...

Homework Statement Three identical capacitors are connected with a resistor in two different ways. When they are connected as in part A of the drawing, the time constant to charge up this circuit is .34s. What is the time constant when they are connected with the same resistor as in part B...

It all makes sense now. Once again thanks for the help.

20= (.008*1680)+(.008*R) R=820 Thank you for all your help!

ok this is what I have so far. 20V= (I*1680)+(I*R) 30V=(I*2930)+(I*R) (I*R)= 20 - (I*1680) (I*R)= 30 - (I*2930) (I*R)=(I*R) 20 - (I*1680)=30 - (I*2930) I=125 But then when I plug 125 back into the two equations to solve for R, I get different values for R in each equations. Looking at the...

I don't understand what your doing with the voltages? I thought I needed to find the current and the resistance of the coil?

How would I start to find either the current or Resistance of the coil because in the equation I would have both of those as unknowns: V= I/R+Rc

Homework Statement Two scales on a voltmeter measure voltages up to 20.0 and 30.0V, respectively. The resistance connected in series with the galvanometer is 1680 ohms for the 20.0V scale and 2930 ohms for the 30.0V scale. Determine the coil resistance and the full-scale current of the...

So using the equation without the current: P=V^2 / R P1=V^2/(R=r) So according to the example vk6kro suggested: V=12 R=15 P=12^2 / 15 P=9.6 (9.6)(.1)=.96 9.6-.96= 8.64=P1 P1= V^2/(R+r) 8.64 = 144/(9.6+r) r=7.07 r/R=7.07/15=.471 I still don't understand how you can find the ratio algebraically...

so: Rs=R+r P1=I^2 Rs and P=I^2 R I'm pretty sure I've understood everything you have said physics wise regarding the problem, but I don't understand how to come up with the answer when the only number I know is 10%. The only difference I see when comparing the two P's is that in P1 I^2 is...

1. P=I^2 R 2. 1/Rp = 1/R + 1/r 3. to find the power dissipated in situation 2 I would use the equation: P=I^2 Rp because they are parallel resistors. So the difference between the to powers is one is multiplied by R and the other by Rp. If I did do everything right I don't understand how...