Recent content by AcidicVision

A proton high above the equator approaches the Earth moving straight downward with a speed of 355 m/s. Find the acceleration of the proton, given that the magnetic field at its altitude is 4.05 X 10^-5 T. Homework Equations F = MA =eVBsin? a = evB/m The Attempt at a Solution...

0.25 would be max and -0.25 would be min with an origin point in the middle?

so I start with cos(110 * t). But I don't have a value for t. Do I estrapolate it from something else?

Homework Statement A vibrating strutural beam in a spacecraft can cause problems if the frequency of vibration is fairly high. Even if the amplitude of vibration is only a fraction of a millimeter, the acceleration can be several times greater than acceleration due to gravity. As an...

I think I resolved this as 62km/h. Thanks chuck. I used t=24/22 = 1.1 - .8421 = .2579 16/.2579 = 62km/h. thanks all for help.

as = distance / elapsed time. sorry was looking at too many things at once.

but isn't average speed found by time/distance? I see the logic in taking more than one hour, not less. But I don't understand exactly how I messed that up. I used the formula for average speed from my book and that was the output.

Homework Statement The position of a particle as a function of time is given by x = ( -2.00 m/s)t + ( 3.00 m/s^3)t^3. (a) Plot x versus t for time from t = 0 to t = 1.00 s. (b) Find the average velocity of the particle from t = 0.150s. to t = 0.250s. (c) find the average velocity from t =...

Im having a rough time with this set of problems in general, so instead of creating a lot of threads I am just going to keep adding what I have here. I am not looking for the answers to my homework, just some guidance and my mistakes pointed out. I appreciate any and all input I receive...