Let F be the force acting on the bowling ball.
Work done by force= change in KE
Let F be the force acting on the bowling ball
Work done by force = change in KE
F .d = 0.5 m V^2 + 0.5 x I x w^2
F d = 0.5 m ( 1+l) V^2
Solve accordingly to get the answer.
Answer: The gravitational force Fg exerted on the orbit by the planet is Fg = G 4/3πr3rhom/ (R1 + d+ R2)^2
Explanation:
Gravitational Force Fg = GMm/r2----1
Where G is gravitational constant
M Mass of the planet, m mass of the orbit and r is the distance between the masses.
Since the circular orbit move around the planet, it means they do not touch each other.
The distance between two points on the circumference of the two massesb is given by d, while the distance from the radius of each mass to the circumferences are R1 and R2 from the question.
Total distance r= (R1 + d + R2)^2---2
Recall, density rho =
Mass M/Volume V
Hence, mass of planet = rho × V
But volume of a sphere is 4/3πr3
Therefore,
Mass M of planet = rho × 4/3πr3
=4/3πr3rho in kg
From equation 1 and 2
Fg = G 4/3πr3rhom/ (R1 + d+ R2)^2
Your answer is C)
a)t=2.78 sec
b)R=835.03 m
c)
Explanation:
Given that
h= 38 m
u=300 m/s
here given that
The finally y=0
So
t=2.78 sec
The horizontal distance,R
R= u x t
R=300 x 2.78
R=835.03 m
The vertical component of velocity before the strike
It is 92.96 millions miles away
Hope that helped :)
