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3 years ago

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Mary and her younger brother Alex decide to ride the 26 ft diameter carousel at the State Fair. Mary sits on one of the horses i

n the outer section at a distance of 11.5 feet from the center. Alex decides to play it safe and chooses to sit in the inner section at a distance of 6 feet from the center. What is Mary's centripetal acceleration aM compared to that of Alex's centripetal acceleration aA?

1 answer:

hammer [34]3 years ago

8 0

Answer:

$a_M=1.92a_A$

Explanation:

$\omega_M=\omega_A$ = Angular speed

$r_M$ = Distance of Mary = 11.5 ft

$r_A$ = Distance of Alex = 6 ft

Ratio of centripetal acceleration is given by

$\dfrac{a_M}{a_A}=\dfrac{\omega_M^2r_M}{\omega_A^2r_A}\\\Rightarrow \dfrac{a_M}{a_A}=\dfrac{r_M}{r_A}\\\Rightarrow a_M=a_A\dfrac{r_M}{r_A}\\\Rightarrow a_M=\dfrac{11.5}{6}a_A\\\Rightarrow a_M=1.92a_A$

Mary's centripetal acceleration is 1.92 times the centripetal acceleration of Alex

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So, the time needed before you hear the splash is approximately <u>2.06 s</u>.

<h3>Introduction</h3>

Hi ! In this question, I will help you. This question uses two principles, namely the time for an object to fall freely and the time for sound to propagate through air. When moving in free fall, the time required can be calculated by the following equation:

$\sf{h = \frac{1}{2} \cdot g \cdot t^2}$

$\sf{\frac{2 \cdot h}{g} = t^2}$

$\boxed{\sf{\bold{t = \sqrt{\frac{2 \cdot h}{g}}}}}$

With the following condition :

t = interval of the time (s)
h = height or any other displacement at vertical line (m)
g = acceleration of the gravity (m/s²)

Meanwhile, for sound propagation (without sound reflection), time propagates is the same as the quotient of distance by time. Or it can be formulated by :

$\boxed{\sf{\bold{t = \frac{s}{v}}}}$

With the following condition :

t = interval of the time (s)
s = shift or displacement (m)
v = velocity (m/s)

<h3>Problem Solving</h3>

We know that :

h = height or any other displacement at vertical line = 19.6 m
g = acceleration of the gravity = 9.8 m/s²
v = velocity = 343 m/s

What was asked :

$\sf{\sum t}$ = ... s

Step by step :

Find the time when the object falls freely until it hits the water. Save value as $\sf{\bold{t_1}}$

$\sf{t_1 = \sqrt{\frac{2 \cdot h}{g}}}$

$\sf{t_1 = \sqrt{\frac{2 \cdot \cancel{19.6} \:_2}{\cancel{9.8}}}}$

$\sf{t_1 = \sqrt{4}}$

$\sf{\bold{t_1 = 2 \: s}}$

Find the time when the sound propagate through air. Save value as $\sf{\bold{t_2}}$

$\sf{t_2 = \frac{h}{v}}$

$\sf{t_2 = \frac{19.6}{343}}$

$\sf{\bold{t_2 \approx 0.06 \: s}}$

Find the total time $\sf{\bold{\sum t}}$

$\sf{\sum t = t_1 + t_2}$

$\sf{\sum t \approx 2 + 0.06}$

$\boxed{\sf{\sum t \approx 2.06}}$

<h3>Conclusion</h3>

So, the time needed before you hear the splash is approximately 2.06 s.

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