As a liquid is added to a beaker, the pressure exerted by the liquid on the bottom
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Answer:
torque τ = 24 N-m
torque τ = 11.98 N-m
Explanation:
given data
mass of the ball = 1.5 kg
arm length l = 70 cm = 0.7 m
mass of arm = 4 kg
solution
we get here center of mass that is
center of mass = ..........1
center of mass =
center of mass = 0.44545 m
and
torque will be
torque τ = F × r ×sinθ ...........2
torque τ = (4+1.5)9.8 × 0.4454 × sin90
torque τ = 24 N-m
and now we get magnitude of the torque
magnitude of the torque τ = F × r ×sinθ
magnitude of the torque τ = (4+1.5)9.8 × 0.4454 × sin60
magnitude of the torque τ = 11.98 N-m
By Newton's second law, the apparent weight of rider coming at rest is <u>3548 N</u>.
According to formula, v = u + at
a = 39.2 m/
Force, F = ma
F = m(a+g)
F = apparent weight
m = mass, given = 65 kg
a = acceleration, given = 39.2 m/
g = 9.8 m/
Put these values in formula, F = m(a+g)
F = 65(39.2+9.8)
F = 3548 N
Therefore, the apparent weight of rider coming at rest is <u>3548 N</u>.
Learn more about apparent weight here:- brainly.com/question/24897276
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