Answer with Explanation:
Let rest mass at point P at distance x from center of the planet, along a line connecting the centers of planet and the moon.
Mass of moon=m
Distance between the center of moon and center of planet=D
Mass of planet=M
We are given that net force on an object will be zero
a.We have to derive an expression for x in terms of m, M and D.
We know that gravitational force=
Distance of P from moon=D-x
=Force applied on rest mass due to m
=Force on rest mass due to mas M
because net force is equal to 0.
Let
Then,
b.We have to find the ratio R of the mass of the mass of the planet to the mass of the moon when x=
Net force is zero
Hence, the ratio R of the mass of the planet to the mass of the moon=4:1
Answer:
The acceleration is - 3 m/s2.
Explanation:
initial speed, u = 54 km/h = 15 m/s
final speed, v = 0
time, t = 5 s
Let the acceleration is a.
use first equation of motion
v = u + at
0 = 15 + a x 5
a = - 3 m/s2
hmax = 5740.48 m. The maximum height that a cannonball fired at 420 m/s at a 53.0° angles is 5740.48m.
This is an example of parabolic launch. A cannonball is fired on flat ground at 420 m/s at a 53.0° angle and we have to calculate the maximum height that it reach.
V₀ = 420m/s and θ₀ = 53.0°
So, when the cannonball is fired it has horizontal and vertical components:
V₀ₓ = V₀ cos θ₀ = (420m/s)(cos 53°) = 252.76 m/s
V₀y = V₀ cos θ₀ = (420m/s)(cos 53°) = 335.43m/s
When the cannoball reach the maximum height the vertical velocity component is zero, that happens in a tₐ time:
Vy = V₀y - g tₐ = 0
tₐ = V₀y/g
tₐ = (335.43m/s)/(9.8m/s²) = 34.23s
Then, the maximum height is reached in the instant tₐ = 34.23s:
h = V₀y tₐ - 1/2g tₐ²
hmax = (335.43m/s)(34.23s)-1/2(9.8m/s²)(34.23s)²
hmax = 11481.77m - 5741.29m
hmax = 5740.48m
With the values you've given, only velocity can be found.
Acceleration is rate of change of velocity
d= 250s
t= 17s
a= d/t
=
= 4.7