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valkas [14]
3 years ago
7

A visitor to a lighthouse wishes to determine the height of the tower. She ties a spool of thread to a small rock to make a simp

le pendulum, which she hangs down the center of a spiral staircase of the tower. The period of oscillation is 10.1 s. The acceleration of gravity is 9.8 m/s 2 . What is the height of the tower?
Physics
1 answer:
masha68 [24]3 years ago
5 0

To solve this problem we must rely on the equations of the simple harmonic movement that define the period as a function of length and gravity as

T = 2\pi \sqrt{\frac{l}{g}}

Where

l = Length

g = Gravity

Re-arrange to find L,

L = g (\frac{T}{2\pi})^2

Our values are given as

g = 9.81m/s

T = 10.1s

Replacing,

L = g (\frac{T}{2\pi})^2

L = (9.81) (\frac{10.1}{2\pi})^2

L = 25.348m

Therefore the height would be 25.348m

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a) v = 21.34 m/s

b) v = 21.34 m/s

c) v = 21.34 m/s

Explanation:

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θ = 26°

The speed of the slow ball as it reaches the ground, v = ?

The initial Kinetic energy of the snow ball, KE_{0}  = 0.5 mu^{2}

Potential energy of the snow ball at the given height, PE = mgh

Final Kinetic energy of the ball as it reaches the ground, KE_{f} = 0.5mv^{2}

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KE_{0} + PE = KE_{f} \\0.5mu^{2} + mgh = 0.5mv^{2}\\v^{2} =2( 0.5u^{2} + gh)\\v^{2} =u^{2} + 2gh\\v = \sqrt{u^{2} + 2gh} \\v = \sqrt{13.3^{2} + 2*9.8*14.2}\\v = 21.34 m/s

b) The speed remains v = 21.34 m/s since it is not a function of the angle of launch

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v = 21. 34 m/s

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