Around midcourse, a velocity adjustment is performed to eliminate the small errors introduced when departing from Earth's orbit. This adjustment is performed using one of the onboard thrusters. At the location where the adjustment is made, the velocity is 26248 m/s and should be 27000 m/s. Knowing that the thruster used for the maneuver generates a thrust of F= 7540 N, determine how long, in minutes, it should be turned on to adjust the velocity. The mass of the spacecraft is 2000 kg.

Explanation:

Given:

Initial velocity, vi = 26248 m/s

Final velocity, vf = 27000 m/s

Force, f = 7540 N

Mass, m = 2000 kg

Force = mass × acceleration

= mass × change in velocity/time

Change in velocity = vf - vi

Time, t = (2000 × (27000 - 26248))/7540

= 199.47 s.

I min = 60 sec

Tine, t = 3.325 min

= 3.33 min.

power out to power in less than 100%

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.