AznBoi said:
66. \varepsilon = Blv \qquad \varepsilon=\frac{-\Delta\phi_{m}}{\Delta t} \qquad \phi_{m}=BAcos\Theta=\;\pi r^{2}\qquad 4\pi b^{2} That's why I chose (e) for 66.
33. I have no idea on this one. I just picked the one that had an atomic number and mass number lower than the beginning nucleus.
38. I tried using mgh=1/2kx^2 . I found (k) first by using F=ma on the mass in equilibrium but ended up with a huge number. Is my method correct?
40. I have no idea how to combine KE with the gravitational force. Please help, Thanks!
\varepsilon=\frac{-\Delta\phi_{m}}{\Delta t} is correct, but which variable (B,\,A,\,\theta) is changing?
\phi_{m}=BAcos\Theta=\;\pi r^{2} does this equality makes sense?
The magnetic field is confined to a < b, and we now that the inductance in loop of radius b is \epsilon, so what of the larger loop?
33. In beta decay, the mass number (A, number of nucleons) remains the same, but the charge of the nucleus (Z) must increase by 1 to (Z+1), because a neutron decays to a proton, electron and anti-neutrino.
38. One can determine the k from the 3 kg mass, which is at (static) equilibrium. F = kx. The 4 kg mass is released, so in addition to its weight, there is a brief acceleration until the spring force exceeds the weight and restores the system. One does apply conservation of energy in this situation.
40. One can use the balance of gravitational force with centripetal force to determine the orbital velocity from which one can determine the kinetic energy.
http://library.thinkquest.org/03oct/02144/basics/orbittal.htm
For a low orbiting satellite, the altitude is much less than the radius of the earth.
See also - Mathematics of Satellite Motion
http://www.glenbrook.k12.il.us/gbssci/phys/Class/circles/u6l4c.html
and also - Lesson 4: Planetary and Satellite Motion
Energy Relationships for Satellites
http://www.glenbrook.k12.il.us/gbssci/phys/Class/circles/u6l4e.html