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Alex787 [66]
3 years ago
15

Some passengers on an ocean cruise may suffer from motion sickness as the ship rocks back and forth on the waves. At one positio

n on the ship, passengers experience a vertical motion of amplitude 1.4 m with a period of 14 s .
A. What is the maximum acceleration of the passengers during this motion?B. What fraction is this of g?
Physics
1 answer:
Evgen [1.6K]3 years ago
4 0

Answer:

0.28198 m/s²

0.02874

Explanation:

A = Amplitude = 1.4 m

T = Time period = 14 s

g = Acceleration due to gravity = 9.81 m/s²

Acceleration is given by

a=\omega^2A\\\Rightarrow a=\left(\frac{2\pi}{T}\right)^2A\\\Rightarrow a=\left(\frac{2\pi}{14}\right)^2\times 1.4\\\Rightarrow a=0.28198\ m/s^2

The maximum acceleration of the passengers during this motion is 0.28198 m/s²

Dividing this value by g

\frac{a}{g}=\frac{0.28198}{9.81}\\\Rightarrow a=0.02874g

The fraction of g is 0.02874

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Answer:

<em>7 kgm/s or 7 Ns</em>

Explanation:

Since the velocity are perpendicular to each other, the impulses will also be perpendicular.

From newton's second law of motion,

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Given: m = 0.14 kg, v₁ = 40 m/s

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I₁ = 5.6 kgm/s

Also for the vertical,

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Substitute into equation 3

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Using Pythagoras to get the resultant impulse

I = √(I₁²+I₂²)................ Equation 4

Substitute the value of I₁ and I₂ into equation 4

I = √(5.6²+4.2²)

I = √(31.36+17.64)

I = √49

<em>I = 7 kgm/s or 7 Ns</em>

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