Answer:
Potential energy
Explanation:
Before release, the catapult has potential energy stored in a tension of torsion device in it. Normally a flexible bow like object that could be made of wood or of metal.
Answer:
a) v = 19,149.6 m/s
b) f = 95%
c) t = 346.5min
Explanation:
First put all values in metric units:

The equation of motion you need is:
where
is the final velocity, a is acceleration and t is time in hours.
Since the spaceship starts from 0 velocity:

Next, you need to calculate the distances traveled on each interval, considering that both starting and final intervals travel the same distance because the acceleration and time are equal. For this part you need the next motion equation:

solving for first and last interval:
Since the spaceship starts and finish with 0 velocity:

Then the ship traveled
at constant speed, which means that it traveled:

Which in percentage is 95% of the trip.
to calculate total time you need to calculate the time used during constant speed:

That added to the other interval times:

A force can be considered a push or pull
hope this helps :)
Answer:
The two forces acting on a boat or some other floating object are buoyancy and gravity
Answer:
i) No, the spring scale does not read a different value
ii) The torque will read a different value, it will reduce
iii) The spring scale does not need to be measured at the center of mass location.
Explanation:
The torque caused by the gyroscope can be given by the relation,
r × f

The torque measured by the gyroscope varies directly with the distance, r.
A decrease in the distance r will also cause a decrease in the value of the torque measured. When the distance, r is reduced from 7.5 inches to 5 inches, the torque caused by the gyroscope's weight also reduces.
The weight of the gyroscope remains constant despite the reduction in the distance because the weight of the gyroscope is not a function of the distance from the gyroscope. Therefore, the spring scale will not read a different value.
Yes, the spring scale does not need to be measured from the center of mass location because the weight does not depend on the location of measurement. The reading of the sprig scale remains constant.