Momentum p = mv is conserved.
before: p = 2.5 * 7.5 + 70 * 0
after: p = (2.5 + 70) * v
p before = p after
2.5 * 7.5 = (2.5 + 70) * v
v = 2.5 * 7.0 / (2.5 + 70)
If the car moves along the distance it will be 16 of the line graph where is independent of the graph
The ball moves by uniformly accelerated motion, and its vertical position at time t is described by the following law
where
is the initial height from which the ball starts its motion
is the initial velocity of the ball
is the gravitational acceleration
The time in which the ball reaches the ground is the time t at which the vertical position y(t) becomes zero:
Which means
whose solutions are:
Neglecting the negative solution (since it has no physical meaning), we can say that the ball reaches the ground after 1.10 s.
Answer:
r = 4.24x10⁴ km.
Explanation:
To find the radius of such an orbit we need to use Kepler's third law:
<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km. </em>
From equation (1), r₁ is:
Therefore, the radius of such an orbit is 4.24x10⁴ km.
I hope it helps you!