Recent content by clamatoman

Volume of the sphere = 4/3 pi r^3 = 2.14×10^9 m^3 Volume of the cone = pi r^2 h/3 = 0.047 m^3 ratio = 2.196 * 10^-11 2.196 * 10^-11 * 2.77*10^20 =6.1*10^9 protons! Thanks!

Ah, I need to consider a sphere, and a conical section of that sphere with the circular base being the 7.5mm pupil. I shall attempt.

Homework Statement (a)Estimate the number of photons per second emitted by a 100-W lightbulb, assuming a photon wavelength of 550nm.(b) A person can just see this bulb for a distance of 800m, with the pupil dilated to 7.5mm. How many photons per second are entering the pupil? Homework...

Well, apparently its 700nm. And 1400 nm for the second order.

Homework Statement Find the smallest grating spacing that let's you see the entire visible spectrum. 400nm to 700nm comprises the visible light spectrum. no other information is given. Homework Equations nλ=d sin θ The Attempt at a Solution I am not sure how to start as all i have been given...

Awesome and i plugged it into make sure the time gives me the same Voltage. Looks right to me. Thanks!

ok using V=Vocos(ωt) rearrange and solve t=(cos^-1(6.36/9))/(1/(√1.5 * 4700*10-6)) = 0.066 seconds? does that look right?

Well in order to solve for the time I need to use charge to use the Q=Qocos(ωt) and rearrange to solve for time. so i worked out the math, my new voltage at 1/2 original energy V=√(2E/C) = 6.36 volts new charge at 1.2 energy Q=CV= (4700e-6 * 6.36) = 0.0299 initial charge Q=CV=(4700e-6 * 9) =...

energy of a capacitor varies with Voltage which is also a factor of charge... so i need to solve for the voltage of the capacitor when energy is 1/2 of initial energy, then plug that into the charge equation, right?

I think I have it. Q=CV=0.0423. ω=1/√LC Q=Qocos(ωt) .5Q=Qocos(ωt) .5=cos(ωt) cos-1(.5)=(ωt) t=(cos-1(.5))/ω t=5.038 seconds Look Good?

so now i just need to solve for time it takes for the capacitor to discharge half of its energy into the inductor...

And solving for the energy i got U=1/2 CV2= ~0.19J, correct?

Let me see if what I am thinking makes sense here for solving current. U=1/2 LI2 for the inductor, U=1/2 CV2 for the capacitor. So does U=U 1/2 CV2 =1/2 LI2 . By doing this and rearrangeing to solve for I i have I=√((CV2)/L) which = ~ 0.5A. IS this correct? or am i way off?

If voltage of the capacitor is zero, then so is voltage of the inductor due to KVL?

when Voltage of the capacitor is 0?