Recent content by Queue

I did convert to square meters. I multiplied by area, didn't divide like I thought I was supposed to which actually makes my units work out. This gives me u = \epsilon_0 E^2 = 8.85*10-12*(3.65*103)2 J/(m3). From there I have u*A = 8.85*10-12*(3.65*103)2*0.000115 J/m. Then I multiplied by...

Homework Statement How much energy is transported across a 1.15 cm2 area per hour by an EM wave whose E field has an rms strength of 36.5 mV/m? Homework Equations u (energy per unit volume) = \epsilon_0 E^2 \frac{J}{m^3} The Attempt at a Solution Since I have u =...

Ok, I have from my book (which was just returned to me) that energy in a wave is given by \epsilon_0 E2 with units of J/m3. So I have u = 8.85*10-12*(36.5*10-3)2. From there I divided by the area I am given (1.15 cm2) and then multiplied by the speed of light (in hours) to give me the right...

That should be \frac{\frac{100 W}{4pi .05^2 m^2}}{c} And for some reason this is coming out as mu_0*8/(2*pi*(.0447213595)) for some reason...

Why can't I simply divide intensity by the speed of light which would make my units work out? I submitted my solutions; Pressure = /frac{/frac{100}{4/pi .05^2}}{c} worked.

How is this not giving me more than I need? The title's radiation is just about light not radiation per se. My units work out if I have (100 W/(4*pi*.05*.05 m2)/c = 1.something*10-5 pascals which are the units I want for the question (thank you masteringphysics)

So am I correct in saying that pressure is intensity over the speed of light? If so, am I also correct in my formula for intensity? I'm not quite sure how I work in momentum/second (momentum/photon and photons/second).

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Homework Statement How much energy is transported across a 1.15 cm^2 area per hour by an EM wave whose E field has an rms strength of 36.5 mV/m? Homework Equations E=hv? The big problem I'm having is that I don't know what formulae to use. The Attempt at a Solution I know what V/m...

Homework Statement Estimate the radiation pressure due to a 100-W bulb at a distance of 5.0 cm from the center of the bulb. Assume that light is complitely absorbed. Estimate the force exerted on your fingertip if you place it at this point. Assume area of the fingertip to be 1.5 cm2...

You're right; I did math poorly (sorry!) and forgot to double the middle term. You have (2000x-1125)2 = 4000000x2-4500000x+1265625 and (2000x+1125)2 = 4000000x2+4500000x+1265625. So (2000x+1125)2 = (2000x-1125)2 + x2 gives us 4000000x2+4500000x+1265625 = 4000000x2-4500000x+1265625 + x2...

Well most of it should cancel out and you should get x2 - 4500000x = 0. Is that what you got? If so what is the answer?

Nor is your initial statement. What notation? For what? With some information perhaps we could be of help.

Well you have (2000x-1125)2+x2 = (2000x+1125)2 and (2000x-1125)2 = 4000000x2-2250000x+1265625 and (2000x+1125)2 = 4000000x2+2250000x+1265625. How does this fall out? Your equation above is incorrect. Further, you don't really need the quadratic equation. Anything of the form ax2 - bx =...

Well you should get (2000x-1125)2+x2 = (2000x+1125)2 from the pythagoreon. Then expand. What did you get from your expansion?