Answer:
By conservation of energy, it can climb up to a height equal to that it went down before. However, due to the friction in the machines, the total mechanical energy of the roller coaster will decrease. As a result, the first "hill" of many roller coasters are the highest, but the followings will have decreasing heights.
Explanation:
Answer:
1. Dheere Dheere (slowly slowly)
2. Har (every)
3. Kal (tomorrow)
4. Mat (don't)
5. Andar (inside)
sorry I wasn't able to write in hindi
the normal force is the force applied opposite to the weight
of the was box. So the normal force is equal to the weight of the box = 24 kg
*(9.81 m/s2) = 235.44 N
the acceleration of the box be solve using newtons 2nd
law of motion:
F = ma
a = F/ m = 585 N/ 24 kg = 24.38 m/s2

Explanation:
Newton's 2nd Law can be expressed in terms of the object's momentum, in this case the expelled exhaust gases, as
(1)
Assuming that the velocity remains constant then

Solving for
we get

Before we plug in the given values, we need to convert them first to their appropriate units:
The thrust <em>F</em><em> </em> is

The exhaust rate dm/dt is


Therefore, the velocity at which the exhaust gases exit the engines is

