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:
2065.005 m/s²
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s²
Now H-h = 42.3 - 0.2 = 42.1 cm = 0.421 m
The final velocity will be the initial velocity
Acceleration of the frog is 2065.005 m/s²