The velocity of a wave with a wavelength of 4.7000 m and frequency of 34.00 hz
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Answer:
The coefficient of friction is (F/(19.6·m)
Explanation:
The given parameters are;
The force applied to the immovable body = F
The time duration the force acts = t
The time the body spends in motion = 3·t
The acceleration due to gravity, g = 9.8 m/s²
From Newton's second law of motion, we have;
The impulse of the force = F × t = m × Δv₁
Where;
Δv₁ = v₁ - 0 = v₁
The impulse applied by the force of friction, is
× (3·t - t) =
× (2·t)
Given that the motion of the object is stopped by the frictional force, we have;
The impulse due to the frictional force = Momentum change = m × Δv₂ = × (2·t)
Where;
Δv₂ = v₂ - 0 = v₂
Given that the velocity, v₂, at the start of the deceleration = The velocity at the point the force ceased to act, v₁, we have;
m × Δv₂ = × (2·t) = m × Δv₁ = F × t
∴ × (2·t) = F × t
= F × t/(2·t) = F/2
The coefficient of dynamic friction, = Frictional force/(The weight of the body) = (F/2)/(9.8 × m)
= (F/(19.6·m)
The coefficient of friction, = (F/(19.6·m)