The correct answer is:
<span>C: in the protons and neutrons of an atom
In fact, the nuclear energy refers to the binding energy of the nucleons (protons and neutrons) of an atom. The protons and the neutrons are held together by the strong nuclear interaction, one of the four fundamental forces of nature, and the energy associated to this interaction is called nuclear energy.
</span>
If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be 
.
<h3>What is an escape velocity?</h3>
The ratio of the object's travel distance over a specific period of time is known as its velocity. As a vector quantity, the velocity requires both the magnitude and the direction. the slowest possible speed at which a body can break out of the gravitational pull of a certain planet or another object.
The formula to calculate the escape velocity of earth is given below:-
Given that earth's mass was half its actual value but its radius stayed the same. The escape velocity will be calculated as below:-
.
Therefore, If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be .
To know more about escape velocity follow
brainly.com/question/14042253
#SPJ4
-- As far as we know, the forces on the wheelbarrow are balanced.
-- That tells us that the net force on the wheelbarrow is zero, just
as if there were no forces acting on it at all.
-- That tells us that the wheelbarrow's acceleration is zero ... its
speed and direction of motion are not changing.
-- That tells us that the wheelbarrow is moving in a straight line
at a constant speed. It's very possible that relative to us, the speed
may be zero, but we can't tell that from the given information.
Answer:
1923 N
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 65 Kg
Radius (r) = 2.5 m
Velocity (v) = 8.6 m/s
Centripetal force (F) =?
The centripetal force, F, can be obtained by using the following formula:
F = mv²/r
F = 65 × 8.6² / 2.5
F = 65 × 73.96 / 2.5
F = 4807.4 / 2.5
F = 1922.96 ≈ 1923 N
Thus, the magnitude of the centripetal's force acting on the student is approximately 1923 N
