The work done by the man pushing the car over the given distance is 1000J.
Given the data in the question;
- Mass of car;

- Acceleration of the car;

- Distance covered by the car;

Work done;
<h3>Work done</h3>
Work done is simply defined as the energy transfer that takes place when an object is either pushed or pulled over a certain distance by an external force. It is expressed as;

Where f is force applied and d is distance travelled.
To determine the work done by the man, we first solve for the force applied F.
From Newton's Second Law; 
We substitute our given values into the expression

Next we substitute our values into the expression of work done above.

Therefore, the work done by the man pushing the car over the given distance is 1000J.
Learn more about work done: brainly.com/question/26115962
Complete Question
Apollo 14 astronaut Alan B. Shepard Jr. used an improvised six-iron to strike two golf balls while on the Fra Mauro region of the moon’s surface, making what some consider the longest golf drive in history. Assume one of the golf balls was struck with initial velocity v0 = 32.75 m/s at an angle θ = 32° above the horizontal. The gravitational acceleration on the moon’s surface is approximately 1/6 that on the earth’s surface. Use a Cartesian coordinate system with the origin at the ball's initial position.
Randomized Variables
vo 32.75 m/s
theta 32 degrees
What horizontal distance, R in meters, did this golf ball travel before returning to the lunar surface?
Answer:
The horizontal distance is
Explanation:
From the question we are told that
The initial velocity is 
The angle is 
The gravitational acceleration of the moon is 
Generally the distance traveled is mathematically represented as

=> 
=>
Answer:
Momentum after = 2400 kgm/s
Explanation:
Momentum after = momentum before
Momentum after = m1v1 +m2v2
Momentum after = (600)(4) + (400)(0)
Momentum after = 2400 kgm/s
Answer:
18.9 m.
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 0 m/s
Final velocity (v) = 70 km/h
Height (h) =?
Next, we shall convert 70 km/h to m/s. This can be obtained as follow:
3.6 km/h = 1 m/s
Therefore,
70 km/h = 70 km/h × 1 m/s / 3.6 km/h
70 km/h = 19.44 m/s
Finally, we shall determine the height. This can be obtained as follow:
Initial velocity (u) = 0 m/s
Final velocity (v) = 19.44 m/s
Acceleration due to gravity (g) = 10 m/s²
Height (h) =?
v² = u² + 2gh
19.44² = 0² + (2 × 10 × h)
377.9136 = 0 + 20h
377.9136 = 20h
Divide both side by 20
h = 377.9136 / 20
h = 18.9 m
Thus, the car will fall from a height of 18.9 m
Answer:0.967meV
Explanation:
-Find the difference in u, so 130.906118-130.90508= 0.001038u
- convert to meV
1 u = 931.494meV
multiply 0.001038 by 931.494
=0.001038 X 931.494
0.967 meV