The acceleration of positive performing SHM is 12cm/sec at distance of 3cm from the mean position its time period is?
1 answer:
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It sounds as though the two people are standing in front of the boat on opposite sides of it, so that they both make an angle of 30.0° with the axis of the boat, as in the attached free body diagram (ignoring the force of buoyancy and the weight of the boat).
By Newton's second law, the net vertical force is
∑ <em>F</em> = <em>P</em>₁ sin(60.0°) + <em>P</em>₂ sin(120.0°) - <em>R</em> = 0
where upward is positive and downward is negative, and the right side is 0 because the boat moves with constant velocity and thus zero acceleration.
We're told that <em>P</em>₁ = <em>P</em>₂ = 600 N, and we know sin(60°) = sin(120°), so the above reduces to
<em>R</em> = 2 <em>P</em> sin(60.0°) = 2 (600 N) sin(60.0°) ≈ 1040 N
Answer:
ωf = 8.8 rad/s
v = 2.2 m/s
Explanation:
We will use the third equation of motion to find the maximum angular velocity of the wheel:
where,
α = angular acceleration = 6 rad/s²
θ = angular displacemnt = 1 rev = 2π rad
ωf = max. final angular velocity = ?
ωi = initial angular velocity = 1.5 rad/s
Therefore,
<u>ωf = 8.8 rad/s</u>
Now, for linear velocity:
v = rω = (0.25 m)(8.8 rad/s)
<u>v = 2.2 m/s</u>