Here is your answer.
what is the magnitude of the compression forces (assumed to be horizontal) acting on both sides of the center board that is sand
wiched between two other boards.Given that the board the board sandwiched between the board weighs 95.5N and if the coefficient of friction between the boards is 0.663
1 answer:
F = normal force by each board on each side
W = weight of the board in between acting in down direction = 95.5 N
f = frictional force in upward direction by each board
= coefficient of friction = 0.663
Using equilibrium of force in Upward direction
f + f = W
f = W/2
f = 95.5/2 = 47.75 N
frictional force is given as
f = F
47.75 = (0.663) F
F = 72.02 N
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Answer:
Thus induced emf is 0.0531 V
Solution:
As per the question:
Diameter of the loop,
Thus the radius of the loop, R = 0.048 m
Time in which the loop is removed, t = 0.15 s
Magnetic field, B = 1.10 T
Now,
The average induced emf, e is given by Lenz Law:
where
= magnetic flux =
where
A = cross sectional area
Also, we know that:
e = 0.0531 V
The sketch is shown in the figure, where I indicates the direction of the induced current.
