Impulse = Ft = (m)(delta v)
delta v = change in velocity = velocity final - velocity initial.
= -22m/s - +18m/s = -40m/s.
mdeltav = (0.40kg)(-40m/s) = -16kgm/s or -16Ns.
Mechanical energy is the answer
Answer:
h f = W + KE
Input energy equals work function plus KE of emitted electron
W = 6.63E-34 * 2.5E15 - 6.3 * 1.6E-19
W = 6.63 * 2.5 * 10^-19 - 10.1 * E-19 ev (1ev = 1.6E-19 J)
W = (16.6 - 10.1)E-19 = 6.5E-19 J
h f = 6.5E-19 J for electrons to be emitted with zero KE
f = 6.5E-19 / 6.63E-34 = .98E-15 / sec = 9.8E-14 / sec (threshold)
If the potential energy of the three-object system is to be a maximum (closest to zero), should object 3 be placed closer to object 1, closer to object 2, or halfway between them?
If the potential energy of the three-object system is to be a maximum (closest to zero), should object 3 be placed closer to object 1, closer to object 2, or halfway between them?
Object 3 should be placed closer to object 1.
Object 3 should be placed on a halfway between object 2 and object 1.
Object 3 should be placed closer to object 2.
Solution
I think that Object 3 should be placed closer to object 2.
