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Tutor profile: Dhananjay T.

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Dhananjay T.
Tutor in Mechanical Engineering .
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Questions

Subject: SAT II Mathematics Level 2

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Question:

Look at this series: 14, 28, 20, 40, 32, 64, ... What number should come next?

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Dhananjay T.
Answer:

This is an alternating multiplication and subtracting series: First, multiply by 2 and then subtract 8. Answer is 56

Subject: Physics (Newtonian Mechanics)

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Question:

A 75.0 kg man stands on a platform scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.20 m/s in 1.00 s. It travels with this constant speed for the next 10.00 s. The elevator then undergoes a uniform acceleration in the negative y direction for 1.70 s and comes to rest. What does the scale register (a) before the elevator starts to move? (b) during the first 1.00 s? (c) while the elevator is traveling at constant speed? (d) during the time it is slowing down? Take g = 10 ms-2.

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Dhananjay T.
Answer:

The scale registers the force applied by it on the man i.e. the normal force. (a) Before the elevator starts to move, the man is in equilibrium. Hence the sum of all forces on him must be zero. Therefore, N = 750 N Hence, the scale registers 750 N before the elevator starts to move. (b) The man accelerates with the elevator at the same acceleration. Applying Newton’s second law in the vertical direction, we get ∑Fy = 750 – N = 75a = 75 X 1.2 which gives N = 660 N Hence, the spring scale registers 660 N for the first 1 s. (c) While the elevator is travelling at constant speed, the acceleration of the man is zero. Therefore, the forces applied on him must add up to zero. which implies N = 750 N (d) The maximum speed of the elevator, v = u + at u = 0 (as the elevator starts from rest) Therefore, v = 1.2 X 1 = 1.2 m/s It then comes to rest in 1.7 s after decelerating uniformly. i.e. 0 = 1.2 + a’t = 1.2 + a’ X 1.7 which gives a’ = -0.7 ms-2 Now, applying Newton’s second law in the vertical plane, we get, ∑Fy = 750 – N = 75a = 75 X (-0.7) which gives N = 802.5 N

Subject: Algebra

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Question:

A man has Rs. 480 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ?

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Dhananjay T.
Answer:

Let number of notes of each denomination be x. Then x + 5x + 10x = 480 16x = 480 x = 30. Hence, total number of notes = 3x = 90.

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