That equation is Newton's universal law of gravitation. ... Any two masses exert equal-and-opposite gravitational forces on each other. If we drop a ball, the Earth exerts a gravitational force on the ball, but the ball exerts a gravitational force of the same magnitude (and in the opposite direction) on the Earth.
Answer:
The light rays falling on a rough surface does follow the laws of reflection. The light rays are incident parallel on the rough surface but due to uneven surface the light rays are not reflected parallel rather they are reflected in different direction. Hence, no image is formed.
To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.
From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

Where,
Angular velocity
v = Lineal Velocity
R = Radius
At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

Where
Angular acceleration
Angular velocity
t = Time
Our values are




Replacing at the previous equation we have that the angular velocity is



Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s
At the same time the angular acceleration would be



Therefore the angular acceleration of a point on the outer edge of the tires is 
Answer:
the impulse experienced by the passenger is 630.47 kg
Explanation:
Given;
initial velocity of the car, u = 0
final velocity of the car, v = 9.41 m/s
time of motion of the car, t = 4.24 s
mass of the passenger in the car, m = 67 kg
The impulse experienced by the passenger is calculated as;
J = ΔP = mv - mu = m(v - u)
= 67(9.41 - 0)
= 67 x 9.41
= 630.47 kg
Therefore, the impulse experienced by the passenger is 630.47 kg
Answer:
The initial velocity of the ball is 28.714 m/s
Explanation:
Given;
time of flight of the ball, t = 2.93 s
acceleration due to gravity, g = 9.8 m/s²
initial velocity of the ball, u = ?
The initial velocity of the ball is given by;
v = u + (-g)t
where;
v is the final speed of the ball at the given time, = 0
g is negative because of upward motion
0 = u -gt
u = gt
u = (9.8 x 2.93)
u = 28.714 m/s
Therefore, the initial velocity of the ball is 28.714 m/s