3 physics questions 請幫忙 t t

1. Using Kepler's Third Law, T^2 is proportional to R^3, where T is the period of revolution of a planet round the sun, and R is the mean distance between the planet and the sun.

Hence, consider the earth, 1^2 = k(Re)^2
where k is a proportional constant, and Re is the earth-sun distance Consider Neptune, 165^2 = k(Rn)^3where Rn is the Neptune-sun distanceThus, [Rn/Re]^3 = 165^2Rn/Re = 165^(2/3) = 30.08
2. c(ii) Let P be the power of the car engine that has been found in Q(b)Energy delivered by car engine = 16.2P
By conservation of energy,
16.2P = (1/2).(1400)(27.8)^2 + 1400gh
where g is the acceleration due to gravity, and h is the vertical height reached.Solve for hd(i) The kinetic energy of the car was dissipated by frictional force between the tyres and the road surface during braking and turned into heat.
(ii) Use: work-done by friction= change of kinetic energy

11200 x 30.5 = (1/2).(1400)v^2
where v is the speed of the car before braking
solve for v gives v = 22.09 m/s = 80 km/h
Thus, John was not telling the truth

3. b(ii) Energy given to the fighter = (1/2).(20000).(80)^2 J = 6.4 x 10^7 J(iii) Power = 6.4x10^7/2 w = 3.2 x 10^6 w

c(i) Energy absorbed by cable = loss of kinetic energy of fighter
= (1/2).(20000).(60)^2 J = 3.6 x 10^7 J
(ii) Average speed of fighter during landing = 60/2 m/s = 30 m/s
Time needed stop the fighter = 80/30 s = 2.667 s
Thus, average power of cable = 3.6x10^7/2.667 w = 1.35 x 10^7 w