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Answer:
A mass is suspended on a spring. The spring is compressed so that the mass is located 5 cm above its rest position. The mass is released at time t = 0 and allowed to oscillate. It is observed that the mass reaches its lowest points after it is released. Find an equation that describes the motion of the mass.
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Explanation:
We do not start at y=0y=0 so it is better to use a cosine curve. The hanging string is compressed to a position that is y=5y=5 units above the rest position. Hence, |a|=5∣a∣=5. Also, since we begin at positive values, then a=5a=5.
We begin at the highest point y=5y=5.
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The angular speed of the merry-go-round is
ω = 0.10 rad/s
The angular moment of inertia of a mass, m, at a radius, r, from the center of the wheel is
I = mr²
Therefore, the angular moment of inertia for the children are
I₁ = (25 kg)*(1.0 m)² = 25 kg-m²
I₂ = (25 kg)*(1.5 m)² = 56.25 kg-m²
The combined angular momentum is
ω(I₁ + I₂) = (0.10 rad/s)*(25 + 56.25 kg-m²)
= 8.125 (kg-m²)/s
Answer: 8.125 (kg-m²)/s
Answer: 19.98m/s
Explanation: since total resistive force (air drag plus rolling friction) for this car has been established to be
100 + 1.2v²] = Fᵈ
The angle of inclination is:
tanθ = 0.061/1 = tan(3.491°)
terminal(constant) speed is achieved when net force down along the incline = Fᵈ
=> mgsin3.491° = [100 + 1.2v²]
=> 768(9.81)0.061 = 100 + 1.2v²
579.62 = 100 + 1.2v²
1.2v² = 579.62 - 100
v² = 479.62/1.2
=> v² = 399.38
=> v ~= 19.98 m/s
So the car achieve terminal speed of 19.98m/m as it's rolling down the grade.