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mixer [17]
3 years ago
8

There are two pendulums with periods 1.5 s and 1 s, which one is longer, and which one is shorter.

Physics
1 answer:
Stells [14]3 years ago
3 0

Answer:

The greater the length, the greater the period. Or vice versa

Explanation:

T = 2π * √(L/g)

Given that T = 1s , g = 9.8m/s^2 , π = 3.14

1 = 2× 3.14 √(L/9.8)

1 = 6.28 ×L /9.8

L= 9.8/6.28

L= 1.56m

When T = 1.5s

1.5 = 2×3.14√(L/9.8)

1.5^2 = 6.28×L/9.8

2.25 = 6.28L/9.8

2.25×9.8 = 6.28L

22.05 = 6.28L

L = 22.05/6.28

L = 3.51m

From the above mathematical solution, it is clear that the greater the length, the greater the period. Or vice versa

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Answer:

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Explanation:

Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.

In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.

By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.

When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.

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