Answer:
Explanation:
Given
At an elevation of , spacecraft is dropping vertically at a speed of
Final velocity of the spacecraft is
using equation of motion i.e.
Insert the values
Therefore, magnitude of acceleration is .
A charge of 2 c is at the origin. when charge q is placed at 2 m along the positive x axis, the electric field at 2 m along the
1 answer:
You might be interested in
Answer:
The work required is -515,872.5 J
Explanation:
Work is defined in physics as the force that is applied to a body to move it from one point to another.
The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.
Kinetic energy (Ec) depends on the mass and speed of the body. This energy is calculated by the expression:
where kinetic energy is measured in Joules (J), mass in kilograms (kg), and velocity in meters per second (m/s).
The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:
In this case:
- W=?
- m= 2,145 kg
- v2= 12
- v1= 25
Replacing:
W= -515,872.5 J
<u><em>The work required is -515,872.5 J</em></u>