Recent content by dmullin4

Thank God for Youtube. I found an equation off a video of Fizeau's experiment on Youtube that gave me the equation: v = 2*d*N*f which is velocity = 2 * distance between mirrors * number of teeth on wheel * frequency Plugging all this in, I got 3.04x10^8 which is pretty darn close to...

Ok, so I have the distance. Where do I get the time and speed? Its a red light, so I know that's a specific wavelength, but will it be similar to the speed of light at 2.99x10^8 m/s?

Here is the information page from the lab experiment. There is nothing mentioned about wavelengths or frequencies, except what I figured out that the wheel needs to turn at 2200 rev/s. Fizeau’s experiment A light source sends a beam of light toward a partially silvered mirror. The...

Homework Statement I know that the light wave must travel 17.3 km or 17300 meters. I found experimentally that the frequency should be 2200 Hz. I picked it from a pool ranging from 0 Hz to 5600 Hz. The goal was to see what frequency was required to get the light to pass correctly through a...

Ok, I have recalculated with r(hat)=1 and got my k values to be very similar to each other. For (-100, 100), I got k=9008, for (-60, 60), I got 8992. Are my numbers off from the known value of k because of the microcoulombs? The known value is 8.99x10^6 and I am getting 8.99x10^3 so it makes...

Homework Statement This is from a lab program for my class. Its set on a grid where I can place charges and use the mouse pointer to find the intensity of the field at any spot. I am using the E values I got from hovering my mouse over the indicated spot (x, y). I have a positive 10...

What do you mean the angle between F1 and F12? My calculations for F12 is F12=k(Q1Q2)/r^2 = 9.0 N*m^2/C^2 * (7.0x10^-6C * 8.0x10^-6C)/(1.20m)^2 = 0.35 N F23 and F31 are similar but with the appropriate charges.

Homework Statement Three charged particles are placed at the corners of an equilateral triangle of side 1.20 m. http://i565.photobucket.com/albums/ss99/dmullin4/HomeworkCH21P13.jpg The charges are +7.0 microcoulombs, -8.0 microcoulombs, and -6.0 microcoulombs. Calculate the magnitude and...

That caculation checks out. Let me try to type it better. {sqrt[1+(19*10^-6)*11]} - 1 = sqrt(1.000209) - 1 = 0.0001044 = 1.044*10^-4

I THINK I GOT IT! Ok, Original Period = 2pi*sqrt(L0/g) at 17 Celsius New Period = 2pi*sqrt(L/g) at 28 Celsius delta T/T0 = T-T0/T0 = (T/T0)-1 [2pi*sqrt(L0/g) / 2pi*sqrt(L/g)] -1 = sqrt(L/L0) -1 = sqrt(L0(1+alpha*delta temperature)/L0) -1 sqrt(1 + alpha*delta temperature) - 1 =...

OK, I got the part where sqrt(L0(1 + alpha * delta T)/L0) would equal sqrt((L0 + L0* alpha * delta T)/T0). What I don't understand is where the 1/2 comes from. Does it have something to do with the original length being 1m and the original period equalling 2s?

Yes, I got that far. So now we use L = L0(1 + alpha * delta T) ? sqrt(L0(1 + alpha * delta T)/L0) = sqrt(1 + alpha * delta T) alpha = 19 x 10-6 delta T = 28 - 17 = 11 sqrt(1 + (19 x 10-6 * 11)) = sqrt(1 + 2.09 x 10-4) sqrt(1.000209) = 1.0001044... So is this number right? If...

Where do you get the information that T = sqrt(L0 + L0*a*t) ?

T/T0 = 2pi * sqrt(L/g) / 2pi * sqrt(L0/g) = sqrt(L/g) / sqrt(L0/g) = sqrt(L) / sqrt(L0) = ? Now I'm lost again.

Ok, so I have T0 = 2pi * sqrt(L0/g) at 17o C and T = 2pi * sqrt(L/g) at 28o Celsius. T is greater than T0 because L is greater than L0. I understand this part. But then they have the clock is losing time by an amount of delta T = T-T0. Where does that come from?