Answer:
1) It expresses the rate (top speed) at which it can move with time.
2) P = 20 W
3) h = 18 km
Explanation:
1) Power is the rate of transfer of energy.
⇒ Power = 
i.e P = 
Thus a car's engine power is 44000W implies that the engine of the car can propel the car at this rate. This expresses the rate (top speed) at which it can move with time.
2) m = 400g = 0.4 kg
t = 20 s
h = 100m
g = 10 m/
P = 
= 
= 
P = 20 W
3) u = 600 m/s
g = 10 m/
From the third equation of free fall,
=
- 2gh
V is the final velocity, U is the initial velocity, h is the height.
0 =
- 2 x 10 x h
0 = 360000 - 20h
20h = 360000
h = 
= 18000
h = 18 km
The maximum height of the bullet would be 18 km.
Answer:
1. You push on the ball and the ball pushes on your hand
.
2. The ball hits the ground and the ground pushes back on the ball
.
3. You walk on the ground with your feet and the ground pushes back on you.
Explanation:
The kinetic energy of a 750 kg car moving at 50.0 km h is 72300 joules.
Given,
mass of car (m)= 750 kg
speed of car (v)= 50 km/h
= 50 x 5/18
= 13.9 m/s
we know that kinetic energy= 1/2 x m x v*v
= 1/2 x 750 x 13.9 *13.9
= 72300 joules
so the kinetic energy is 72300 joules.
Energy is the capacity of an object to do work, and like work, energy's unit is the joule (J). Energy exists in many different forms, but the one we think of most often when we think of energy is kinetic energy. Kinetic energy is often thought of as the energy of motion because it is used to describe objects that are moving.
Learn more about kinetic energy here :-
brainly.com/question/999862
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Okay whats the question though?
Answer:
Maximum force, F = 1809.55 N
Explanation:
Given that,
Diameter of the anterior cruciate ligament, d = 4.8 mm
Radius, r = 2.4 mm
The tensile strength of the anterior cruciate ligament, 
We need to find the maximum force that could be applied to anterior cruciate ligament. We know that the unit of tensile strength is Pa. It must be a type of pressure. So,

So, the maximum force that could be applied to anterior cruciate ligament is 1809.55 N