Thank you for visiting A parent exerts a torque on a merry go round at a park The torque has a magnitude given by tex 2 50 times 10. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
a) Work done is 2.679 * 10^(-12) Nm.
b) Power is 0.3348 * 10^(-12) W
The work done by the torque for one complete cycle can be found by multiplying the torque with the angular displacement. The power is then found by dividing the work done by the time taken.
The subject of this question is Physics, specifically the concepts of work, power and torque related to a rotating object such as a merry-go-round.
In general, the work done by a force (in this case provided by a torque) is the product of the force, the displacement it covers and the cosine of the angle between the force direction and displacement direction. However, when it comes to a rotating object, the displacement here is considered as an angular displacement. The work done becomes the product of the torque (τ) and the angle (θ) through which it acts (in radians).
Firstly, to determine how much work (Work) is performed by the torque as the merry go round rotates through 1.70 revolutions, we need to convert the revolutions to radians (as 1 revolution equals to 2π radians). Multiply the result by the given torque.
Secondly, to find the power (Power) transferred by the parent to the ride, we use the formula Power = Work / time. Power here is the rate at which work is done or energy is transferred. In this case, the time taken for the 1.70 revolutions is given as 8.00s.
Work= 10.693 * (2.507 * 10^(-13)) = 2.679 * 10^(-12) Nm.
Work= 2.679 * 10^(-12) Nm.
Power= 2.679 * 10^(-12)/8.00
Power= 0.3348 * 10^(-12) W
Hence, power transferred by the parent to the ride is 0.3348 * 10^(-12) W.
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