1)
The capacitance of a parallel-plate capacitor is given by:
where
is the vacuum permittivity
A is the area of each plate
d is the distance between the plates
Here, the radius of each plate is
so the area is
While the separation between the plates is
So the capacitance is
And now we can find the energy stored,which is given by:
2) 0.71 J/m^3
The magnitude of the electric field is given by
and the energy density of the electric field is given by
and using
, we find
Hi there!
Use the equation:
Where m2 and v2 deal with the larger object, and m1 and v1 with the smaller object. Plug in the given values:
v2 = ?
m1 = 0.048 kg (converted)
m2 = 2.95
v1 = 391
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:
Replacing:
W= -515,872.5 J
<u><em>The work required is -515,872.5 J</em></u>