Recent content by myersb05

The light beam shown in Figure P22.21 makes an angle of ϕ = 23.5° with the normal line NN' in the linseed oil. Determine the angles θ and θ'. (The refractive index for linseed oil is 1.48.) The normal line makes a 90 angle with the surface of the linseed oil. The light then passes through...

(SOLVED) Calculator Entry ErrorAngles of Reflection of Light A ray of light strikes a flat 2.00 cm thick block of fused quartz (n = 1.458) at an angle of θ = 27.0° with the normal (Fig. P22.18). Trace the light beam through the fused quartz and find the angles of incidence and refraction at...

Yea, I'll have to remember that

Got it, thanks again Al

Yea, that is exactly what I typed into my TI-89 and I got exactly .000004

WebAssign is telling me that .000004 is close but that I should keep more digits. I can't come up with an answer that is any less rounded than .000004.

Worked perfectly. Thanks again Al. That was the same problem I had last night. My equation was written improperly in my notes. I am stuck on another one if you have time. Two long parallel conductors separated by 14.0 cm carry currents in the same direction. The first wire carries a...

A cardiac pacemaker can be affected by a static magnetic field as small as 1.7 mT. How close can a pacemaker wearer come to a long, straight wire carrying 27 A? I know that F=ILBsin(theta). I attempted to solve using I=27, L as unknown, F=1.7, and Bwire=mupi*I/2pir with an unknown r that...

I felt like this problem was simple. We submit our answers in WebAssign however, and I can not find the correct solution. Two parallel wires are 20.0 cm apart, and each carries a current of 40.0 A. (a) If the currents are in the same direction, find the force per unit length exerted on...

Oh wait! You can solve that because the v's eventually cancel out. Thanks al, you are the man. You helped me maintain my 100 homework average for summer physics 2!

When solving qvB=m(v^2/R) for R, I got R=mv^2/qvB and from what I said earlier, I know that 2piR/V=T. So I can solve that for R as well and get R=TV/2pi. Then I can set my two equations equal and I have mv^2/qvB=TV/2pi. So now I successfully have time in my equation. Thanks so much for the...

Were you referring to the fact that F would also have to equal m*(v^2/R)

I had tried to set F=0 because the proton has a constant speed and no acceleration. But that did not help because you can't solve 0=qvB for B which is what I need. R is related to the time it takes for a revolution by 2piR/V. However, I have no radius. If I am wrong about F being equal to...

A proton moving freely in a circular path perpendicular to a constant magnetic field takes 1.50 µs to complete one revolution. Determine the magnitude of the magnetic field. I know that a proton has a charge of 1.6x10^-19 and a mass of 1.67x10^-27. Those are the only two relevant facts I...