Potential energy is stored energy. For example, if a bowling ball is on top of a giant hill, we say it has potential energy because it has the potential to do work which is to roll down the hill.
Kinetic energy is the energy of movement so once that ball rolls down that hill, that potential energy is converted to kinetic energy.
<span>Shading.
When light hits an opaque surface some is absorbed, the rest is reflected, The reflected light is called shading. Reflection is not simple and varies with material.
The surface’s structure defines the details of reflection. Variations produce anything from bright specular reflection</span>
(a) 0.448
The gravitational potential energy of a satellite in orbit is given by:
where
G is the gravitational constant
M is the Earth's mass
m is the satellite's mass
r is the distance of the satellite from the Earth's centre, which is sum of the Earth's radius (R) and the altitude of the satellite (h):
r = R + h
We can therefore write the ratio between the potentially energy of satellite B to that of satellite A as
and so, substituting:
We find
(b) 0.448
The kinetic energy of a satellite in orbit around the Earth is given by
So, the ratio between the two kinetic energies is
Which is exactly identical to the ratio of the potential energies. Therefore, this ratio is also equal to 0.448.
(c) B
The total energy of a satellite is given by the sum of the potential energy and the kinetic energy:
For satellite A, we have
For satellite B, we have
So, satellite B has the greater total energy (since the energy is negative).
(d) 
The difference between the energy of the two satellites is:
As the ball is moving in air as well as we have to neglect the friction force on it
So we can say that ball is having only one force on it that is gravitational force
So the force on the ball must have to be represented by gravitational force and that must be vertically downwards
So the correct FBD will contain only one force and that force must be vertically downwards
So here correct answer must be
<em>Diagram A shows a box with a downward arrow. </em>
Answer:
a) t₁ = 4.76 s, t₂ = 85.2 s
b) v = 209 ft/s
Explanation:
Constant acceleration equations:
x = x₀ + v₀ t + ½ at²
v = at + v₀
where x is final position,
x₀ is initial position,
v₀ is initial velocity,
a is acceleration,
and t is time.
When the engine is on and the sled is accelerating:
x₀ = 0 ft
v₀ = 0 ft/s
a = 44 ft/s²
t = t₁
So:
x = 22 t₁²
v = 44 t₁
When the engine is off and the sled is coasting:
x = 18350 ft
x₀ = 22 t₁²
v₀ = 44 t₁
a = 0 ft/s²
t = t₂
So:
18350 = 22 t₁² + (44 t₁) t₂
Given that t₁ + t₂ = 90:
18350 = 22 t₁² + (44 t₁) (90 − t₁)
Now we can solve for t₁:
18350 = 22 t₁² + 3960 t₁ − 44 t₁²
18350 = 3960 t₁ − 22 t₁²
9175 = 1980 t₁ − 11 t₁²
11 t₁² − 1980 t₁ + 9175 = 0
Using quadratic formula:
t₁ = [ 1980 ± √(1980² - 4(11)(9175)) ] / 22
t₁ = 4.76, 175
Since t₁ can't be greater than 90, t₁ = 4.76 s.
Therefore, t₂ = 85.2 s.
And v = 44 t₁ = 209 ft/s.
