會考物理MC-幫幫忙 ><”

Q.31
f=1/2 mg
mgsin30 + f = F,
F is the force make it upward
work done = Fs =(mgsinA + f)s = (mgsin30 +1/2mg)s
Power = E/t = work done /t
[(mgsin30 +1/2mg)s ]/t = mgs/t ,
(s=vt) mgvt/t = mgv
so, ans is B

Q.32
mgh = 1/2mv^2 + work done against friction
because,no acceleration, so no K.E. gain
so, mgh = work done against friction
mgh= Fs = mgsinA x 2 = work done against friction= 10J

Q.35
it is a similar triangle( maths methods)
(1.5)/(3000 +1.5) = 0.6/(the distance between two clouds)
the distance between two clouds= 1200.6003m
the distance between two clouds/time =1200.6003/ 20 = 60 ms-1

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