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3 years ago

If a particle undergoes shm with amplitude 0.21 m, what is the total distance it travels in one period?

1 answer:

7 0

If a particle undergoes simple harmonic motion with an amplitude of 0.21 meters, this means that the maximum displacement of the particle from its resting position is 0.21. For one period, it traveled from its starting position which is twice the amplitude and then back to its original position which is another distance that is twice the amplitude as well. Therefore, the total distance it traveled is 2*amplitude + 2*amplitude = 2*0.21 + 2*0.21 = 0.42 + 0.42 = 0.84 meters.

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tatuchka [14]

Answer:

7.04 m

Explanation:

t = Time taken

u = Initial velocity

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$v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 1.2-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 1.2}\\\Rightarrow u=4.85\ m/s$

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3 years ago

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2 years ago

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Answer: The correct answer is (b) F1 does more work than F2.

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While F2 is at angle so horizontal component of F2 is F2cos(theta) and as we know cos is a decreasing in 0 to 90 degree. Therefore F1 does more work than F2.

Work is defined as the product of the magnitude of a force and the displacement of the object it is acting upon. In this case, the forces F1 and F2 are acting on the box, causing it to move across the floor. To calculate the work done by each force, we need to calculate the magnitude of the force and the displacement of the box.

The magnitude of F1 is given in the diagram, and it is 2 N. The magnitude of F2 is also given, and it is 4 N. For the displacement of the box, we will assume it is 1 m.

Now we can use the formula W = F x d to calculate the work done by each force.

For F1: W1 = 2 N x 1 m = 2 J

For F2: W2 = 4 N x 1 m = 4 J

Since F1 does 2 J of work, and F2 does 4 J of work, F1 does more work than F2 does. Therefore, the correct answer is (b) F1 does more work than F2 does.

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