The back wall of a home aquarium is a mirror that is a distance of 46.0 cm away from the front wall. The walls of the tank are n
1 answer:
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The electrical force acting on a charge q immersed in an electric field is equal to
where
q is the charge
E is the strength of the electric field
In our problem, the charge is q=2 C, and the force experienced by it is
F=60 N
so we can re-arrange the previous formula to find the intensity of the electric field at the point where the charge is located:
where
q is the charge
E is the strength of the electric field
In our problem, the charge is q=2 C, and the force experienced by it is
F=60 N
so we can re-arrange the previous formula to find the intensity of the electric field at the point where the charge is located:
The magnitude of the friction force is 25 N
Explanation:
To solve this problem, we just have to analyze the forces acting on the block along the horizontal direction. We have:
- The horizontal component of the pulling force,
, where F = 50 N is the magnitude and
is the angle between the direction of the force and the horizontal; this force acts in the forward direction
- The force of friction,
, acting in the backward direction
According to Newton's second law, the net force acting on the block in the horizontal direction must be equal to the product between the mass of the block and its acceleration:
where
m is the mass of the block
is the horizontal acceleration
However, the block is moving at constant speed, so the acceleration is zero:
So the equation becomes
(1)
The net force here is given by
(2)
And so, by combining (1) and (2), we find the magnitude of the friction force:
Learn more about force of friction:
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