Considering the definition of kinetic energy, the bullet has a kinetic energy of 156.25 J.
<h3>Kinetic energy</h3>
Kinetic energy is a form of energy. It is defined as the energy associated with bodies that are in motion and this energy depends on the mass and speed of the body.
Kinetic energy is defined as the amount of work necessary to accelerate a body of a given mass and in a rest position, until it reaches a given speed. Once this point is reached, the amount of accumulated kinetic energy will remain the same unless there is a change in speed or the body returns to its rest state by applying a force to it.
The kinetic energy is represented by the following expression:
Ec= ½ mv²
Where:
- Ec is the kinetic energy, which is measured in Joules (J).
- m is the mass measured in kilograms (kg).
- v is the speed measured in meters over seconds (m/s).
<h3>Kinetic energy of a bullet</h3>
In this case, you know:
Replacing in the definition of kinetic energy:
Ec= ½ ×0.500 kg× (25 m/s)²
Solving:
<u><em>Ec= 156.25 J</em></u>
Finally, the bullet has a kinetic energy of 156.25 J.
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Answer:
The total momentum of the universe is always the same and is equal to zero. The total momentum of an isolated system never changes. Momentum can be transferred from one body to another.
Momentum quantifies how likely an object is to stay in motion. Momentum can also be explained using the equation, p=mv, where p is equal to momentum, m is equal to mass, and v is equal to velocity.
Explanation:
Yes that is correct or in other form, True
D-It will become a temporary magnet because the domains will easily realign.
Answer:
a. 79.1 N
b. 344 J
c. 344 J
d. 0 J
e. 0 J
Explanation:
a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force 
by the worker must be equal to the friction force 
on the crate, which is the product of friction coefficient μ and normal force N:
Let g = 9.81 m/s2
b. The work is done on the crate by this force is the product of its force and the distance traveled s = 4.35
c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35
This work is negative because the friction vector is in the opposite direction with the distance vector
d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.
e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction
