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zzz [600]
4 years ago
12

A physics student with a stopwatch drops a rock into a very deep well, and measures the time between when he drops the rocks and

when he hears the sound of the rock hitting the water below. If the speed of sound is 343 m/s, and the student measures a time of 6.20 s, how deep is the well?
Physics
1 answer:
tankabanditka [31]4 years ago
6 0

Answer:

h= 161.06 m

Explanation:

Given that

Speed of the sound ,C= 343 m/s

Total time ,t= 6.2 s

lets take the depth of the well = h

The time taken by stone before striking the water = t₁

we know that

h=\dfrac{1}{2}gt_1^2

t_1=\sqrt{\dfrac{2h}{g}}

The time taken by sound =t₂

t_2=\dfrac{h}{343}

The total time

t = t₁+ t₂

6.2 = \sqrt{\dfrac{2h}{g}}+\dfrac{h}{343}

6.2 = \sqrt{\dfrac{2h}{9.81}}+\dfrac{h}{343}

Now by solving the above equation we get

h= 161.06 m

Therefore the depth of the well will be 161.06 m.

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Given:

The angular velocity of the disk is 10\,{{{\text{rad}}} \mathord{\left/{\vphantom {{{\text{rad}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}.

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Concept:

The linear speed of the bug present at the rim of the circular disk is given by.

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Here, v is the linear speed,r is the radius of the disk and \omega is the angular speed of the disk.

Substitute 0.1\,{\text{m}} for r and 10\,{{{\text{rad}}} \mathord{\left/{\vphantom {{{\text{rad}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} for \omega in above expression.

\begin{aligned}v &= 0.1\,{\text{m}} \times 10\,{{{\text{rad}}} \mathord{\left/{\vphantom {{{\text{rad}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\&= 1\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} \\\end{aligned}  

The magnitude of the centripetal acceleration of the bug cling to the rim of the disk is given as.

{a_c} = \dfrac{{{v^2}}}{r}  

Substitute 1\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} for v and 0.1\,{\text{m}} for r in above equation.

\begin{aligned}a_c&=\dfrac{(1\text{m/s})^2}{0.1\text{m}}\\&=\frac{1}{0.1}\text{m/s}^2\\&=10\,\text{m/s}^2\end{aligned}  

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Learn More:

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Answer Details:

Grade: College

Chapter: Uniform Circular motion

Subject: Physics

Keywords:  Circular disk, circular motion, angular speed, linear speed, bug clinging, rim of the disk, acceleration, magnitude, constant, rad/s.

7 0
4 years ago
Read 2 more answers
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