To solve this problem we will apply the concepts related to potential gravitational energy. This is defined as the product between mass, acceleration and change in height and can be expressed as,

Here,
m = Mass
g = Gravitational acceleration
= Height
Replacing with our values we have,


Therefore the change in gravitational potential energy is 883J.
The answer to this question would possibly be E.
When this component is achieved when your camera takes a photograph that is a GOAL.
Complete Question
Use Stefan's law to find the intensity of the cosmic background radiation emitted by the fireball of the Big Bang at a temperature of 2.81 K. Remember that Stefan's Law gives the Power (Watts) and Intensity is Power per unit Area (W/m2).
Answer:
The intensity is
Explanation:
From the question we are told that
The temperature is 
Now According to Stefan's law

Where
is the Stefan Boltzmann constant with value 
Now the intensity of the cosmic background radiation emitted according to the unit from the question is mathematically evaluated as

=> 
=> 
substituting values


First we have to convert:
75 km / h ( * 3.6 ) = 270 m/s
18 m/s = 64.8 m/s
d = v i · t + 1/2 a · t²
v = v i + a t
-----------------------------------
4000 = 64.8 · t + 1/2 a · t²
270 = 64.8 + a t
a = ( 270 - 64.8 ) / t
a = 205.2 / t
4000 = 64.8 t + 1/2 · ( 205.2 / t ) · t²
4000 = 64.8 t + 102.6 t
4000 = 167.4 t
t = 4000 : 167.4 = 23.89 s
a = 205.2 : 23.89 = 8.58 m/s²
Answer: His average acceleration was 8.58 m/s²