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Maurinko [17]
2 years ago
12

A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A cons

tant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m
Physics
1 answer:
Virty [35]2 years ago
8 0

Complete Question

A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.What will be the equilibrium height of the mass?

Answer:

H_m=1.65m

H_E=1.16307m

Explanation:

From the question we are told that

Mass of ball M=2kg

Length of string L= 2m

Wind force F=13.2N

Generally the equation for \angle \theta is mathematically given as

tan\theta=\frac{F}{mg}

\theta=tan^-^1\frac{F}{mg}

\theta=tan^-^1\frac{13.2}{2*2}

\theta=73.14\textdegree

Max angle =2*\theta= 2*73.14=>146.28\textdegree

Generally the equation for max Height H_m is mathematically given as

H_m=L(1-cos146.28)

H_m=0.9(1+0.8318)

H_m=1.65m

Generally the equation for Equilibrium Height H_E is mathematically given as

H_E=L(1-cos73.14)

H_E=0.9(1+0.2923)

H_E=1.16307m

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a)     L = 33.369 m , b) 21

Explanation:

The analysis of the ocean depth can be performed assuming that at the bottom of the ocean there is a node and the surface must have a belly, so the expression for resonance is

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The speed of the wave is

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Let's solve the two equations

       n₁ 1,589 = n₂ 1,459

       n₁ / n₂ = 1.4539 / 1.589

       n₁ / n2 = 0.91498

Since the two frequencies are very close the whole numbers must be of consecutive resonances, let's test what values ​​give this value

  n₁    n₂    n₁ / n₂

  1      3       0.3

  3     5       0.6

  5     7        0.7

  7     9        0.77

  9    11        0.8

  17   19       0.89

  19  21        0.905

  21  23       0.913

  23 25       0.92

Therefore the relation of the nodes is n₁ = 21  and n₂ = 23

Let's calculate

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                L = 21  1,589

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