Q = mcΔT
<span>q = 55.8g x 0.450J/gC x 23.5C </span>
<span>q = 590. J ................ to three significant digits
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Answer:
50,000 V/m
Explanation:
The electric field between two charged metal plates is uniform.
The relationship between potential difference and electric field strength for a uniform field is given by the equation

where
is the potential difference
E is the magnitude of the electric field
d is the distance between the plates
In this problem, we have:
is the potential difference between the plates
d = 15 mm = 0.015 m is the distance between the plates
Therefore, rearranging the equation we find the strength of the electric field:

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Answer:
-v/2
Explanation:
Given that:
- Collides with the wall going through a sliding motion on on the plane smooth surface.
- Upon rebounding from the wall its kinetic energy becomes one-fourth of the initial kinetic energy before collision.
<u>We know, kinetic energy is given as:</u>

consider this to be the initial kinetic energy of the body.
<u>Now after collision:</u>


Considering that the mass of the body remains constant before and after collision.

Therefore the velocity of the body after collision will become half of the initial velocity but its direction is also reversed which can be denoted by a negative sign.