Answer:
1.25 m
Explanation:
From the question given above, the following data were obtained:
Force ratio = 2.5
Distance of load from the fulcrum = 0.5 m
Distance of effort =.?
The distance of the effort from the fulcrum can be obtained as illustrated below:
Force ratio = Distance of effort / Distance of load
2.5 = Distance of effort / 0.5
Cross multiply
Distance of effort = 2.5 × 0.5
Distance of effort = 1.25 m
Therefore, the distance of the effort from the fulcrum is 1.25 m
Answer:
Option D
Explanation:
The work done can be given by the mechanical energy used to do work, i.e., Kinetic energy and potential energy provided to do the work.
In all the cases, except option D, the energy provided to do the useful work is not zero and hence work done is not zero.
In option D, the box is being pulled with constant velocity, making the acceleration zero and thus Kinetic energy of the system is zero. Hence work done in this case is zero.
The velocity of the body is zero; option A
<h3>What is the motion of an oscillating body?</h3>
The motion of an oscillating body is known as simple harmonic motion.
Simple harmonic motion involves a periodical motion of a body whose acceleration is directed towards a fixed point.
For a body that is oscillating up and down at the end of a spring, considering when the body is at the top of its up-and-down motion, the velocity of the body at the top and down is zero since the body comes to rest at the top and down position of its motion.
In conclusion, oscillating bodies undergo simple harmonic motion.
Learn more about simple harmonic motion at: brainly.com/question/24646514
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Here, height is given which will be the distance for a freely falling object.
The velocity will be
and the acceleration will be
In this way, the formula works.
Periodic time is the time taken for one complete oscillation by a body in circular motion. In this case the merry-go round takes 2 minutes to cover 15 complete oscillations. 2 Minutes = 120 seconds
Hence, 15 oscillations takes 120 secs
thus 1 oscillation takes 120/15 = 8 seconds
therefore the period of the merry-go-round = 8 seconds
