Answer:

Explanation:
As we know that the orbital speed is given as

here we know that
v = 5500 m/s


now we have


now acceleration due to gravity of planet is given as



now range of the projectile on the surface of planet is given as



Hello,
<u>Solution for A:</u>
Force = 3.00N
Mass = 0.50 Kgs
Time = 1.50 Seconds
According to newton's second law of motion;
Force = Mass times Acceleration(a)
3.00 = 0.50 * a
a = 3.00/0.50 = 6.00 m/s^2
We know that acceleration = Velocity / time
So Velocity = time * acceleration = 1.50 * 6 = 9.00 m/s^2
<u>Solution for B:</u>
The net force = 4.00N - 3.00N = 1.00N to the left
Force = 1.00N
Mass = 0.50Kg
Time = 3.00 Seconds
Again; F = MA (Where F is force, M is mass and A is acceleration)
1.00N = 0.5 * A
A = 1/0.5 = 2 m/s^2
Velocity = Acceleration * Time = 2 * 3 = 6 m/s
Time period = time/no. of waves = 6/3 = 2s
Initially, the spring stretches by 3 cm under a force of 15 N. From these data, we can find the value of the spring constant, given by Hook's law:

where F is the force applied, and

is the stretch of the spring with respect to its equilibrium position. Using the data, we find

Now a force of 30 N is applied to the same spring, with constant k=5.0 N/cm. Using again Hook's law, we can find the new stretch of the spring: