29. The earth has a mass of 5.98 x 1024kg and the moon has a mass of 7.35 x 1022kg. The distance from the centre of the moon to the centre of the earth is 3.84 x 108m. A rocket with a total mass of 1200kg is 3.0 x 108m from the centre of the earth and directly in between the earth and the moon. Find the net gravitational force on the rocket from the earth and moon.
If the rocket is 3.0 x 108 m away from the earth, then it is 3.84 x 108 – 3.0 x 108 = 0.84 x 108 m away from the moon.
The net gravitational force on the rocket is 4.48 N towards the earth.
30.A 12 kg meteor…show more content… F=(Gm_1 m_2)/r^2
= (6.67×〖10〗^(-11) )(5.98×〖10〗^24 )(30)/(7.44 ×〖10〗^6 )^2
=216 N
The meteor will experience 216 N of force.
31. A 500kg satellite experiences a gravitational force of 3000N, while moving in a circular orbit around the earth.
a) Find the radius of the circular orbit. F=(Gm_1 m_2)/r^2 r= √((Gm_1 m_2)/F)
= √(((6.67×〖10〗^(-11))(5.98 ×〖10〗^24)(500))/3000)=8.15 ×〖10〗^6 m
The radius is 8.15 x 106 m
b) Find the speed of the satellite. v=√((Gm_E)/r)
=√((6.67×〖10〗^(-11) )(5.98×〖10〗^24 )/(8.15×〖10〗^6 ))=6996 m/s
The speed is 6996 m/s
c) Find the period of the orbit. T=2πr/v
= 2π(8.15×〖10〗^6 )/6996=7320 s
The period of the orbit is 7320 s
32. To simulate gravity, a circular space station with a radius of 150 m is rotated so that astronauts standing on the inner surface move at 30 m/s. If the 75 kg astronaut stands on a bathroom scale, what reading will it give? (Assume that the scale is calibrated in newtons.)
Only the normal force is acting on the astronaut
F_N=F_c
=(mv^2)/r
=((75) (30)^2)/150=450 N
The scale will read 450 N
33. Examine the charge distribution shown.
a) Find the net force on charge 1.
F_E31= (kq_1…show more content… N=q/e
=(4.80×〖10〗^(-19))/(1.6×〖10〗^(-19) )=3
The sphere has a deficit of 3 electrons (since it has a positive charge)
39. A magnetic field of 0.02000T [up] is created in a region.
a) Find the initial magnetic force on an electron initially moving at 5.00 x 106 m/s [N] in the field.
Magnitude of the force:
F_m=qvB
=(1.6×〖10〗^(-19) )(5.00×〖10〗^6 )(0.0200)=1.6×〖10〗^(-14) N
Using right hand rule, when fingers point up and thumb points north, the palm faces east
The initial magnetic force on the electron is 1.6 x 10-14 N [E]
b) What is the radius of the circular path? Make a sketch showing the path of the electron.
F_c=F_m
(mv^2)/r=qvB
r=mv/qB
=(9.11×〖10〗^(-31) )(5.00×〖10〗^6 )/((1.60×〖10〗^(-19))(0.02))=0.00142 m
The radius of the path is 1.42 mm
40. A magnetic field of 1.4 T [N] is 4.0m wide. A very long wire crosses the field.
a) If the current in the wire is 2.0A [W], find the magnetic force on the wire.
Magnitude of the force
F_M=ILB
=(2.0)(4.0)(1.4)=11.2 N
Using right hand rules, if fingers point north and thumb points west, the palm faces downwards.
The magnetic force on the wire is 11.2 N [downwards]
