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vampirchik [111]
3 years ago
14

You decide to visit Santa Claus at the north pole to put in a good word about your splendid behavior throughout the year. While

there, you notice that the elf Sneezy, when hanging from a rope, produces a tension of 445 N in the rope.
If Sneezy hangs from a similar rope while delivering presents at the earth's equator, what will the tension in it be?
Physics
1 answer:
Fynjy0 [20]3 years ago
3 0

Answer:

Explanation:

The question here is that if sneezy hands from a similar rope while delivering presents at the earth's equator, what will be the tension in the rope be. Here is the solution:The tension on the rope when it is at pole, T= 455 NTo find, the tension, t= mgTo solve for mass, m= t/g. Substituting this we have, m=455/9.8. m=46.43 kgAssume that the downwards acceleration is, a= -46.43 m/s^2.T = mg + maT = (46.43 kg) ( 9.8 m/s^2) - (46.43 kg) (-46.43 m/s^2)T = 455.01 kg-m/s^2 - -2155.74 kg-m/s^2T = 2610.75 kg-m/s^2 = 2610.75 N

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Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. He decides to move them at the
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Answer:

Part a)

a= 0.32 m/s^2

Part b)

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Part c)

F_c = 5.5 N

Explanation:

Part a)

As we know that the friction force on two boxes is given as

F_f = \mu m_a g + \mu m_b g

F_f = 0.02(10.6 + 7)9.81

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so we have

F_p - F_f = (m_a + m_b) a

9.1 - 3.45 = (10.6 + 7) a

a = \frac{5.65}{17.6}

a= 0.32 m/s^2

Part b)

For block B we know that net force on it will push it forward with same acceleration so we have

F_c - F_f = m_b a

F_c = \mu m_b g + m_b a

F_c = 0.02(7)(9.8) + 7(0.32)

F_c = 3.6 N

Part c)

If Alex push from other side then also the acceleration will be same

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F_p - F_f - F_c = m_b a

9.1 - 0.02(7)(9.8) - F_c = 7(0.32)

F_c = 9.1 - 0.02 (7)(9.8) - 7(0.32)

F_c = 5.5 N

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Answer:

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Answer:

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