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mihalych1998 [28]
3 years ago
14

A hiker is at the bottom of a canyon facing the canyon wall closest to her. She is 790.5 meters from the wall and the sound of h

er voice travels at 329.2 m/s at that location. How long after she shouts will she hear her echo? Round to two decimals. (Be careful to consider why echoes happen)
Physics
1 answer:
tester [92]3 years ago
7 0

Answer:

4.80 seconds

Explanation:

The velocity of sound is obtained from;

V= 2d/t

Where;

V= velocity of sound = 329.2 ms-1

d= distance from the wall = 790.5 m

t= time = the unknown

t= 2d/V

t= 2 × 790.5/ 329.2

t= 4.80 seconds

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Answer:

\frac{y}{L} = 0.66

Hence, the fraction of the length of the rod above water = \frac{y}{L} = 0.66

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Explanation:

Data given:

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Let's denote density of Water with w

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r = 5/9 x w

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In order to find this, we first need to find out the all type of forces acting upon the rod.

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As, we know

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where,

X (L-y) = volume

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Another force acting upon it is:

F = mg

F2 =  X x L x r x g

Now, the torques acting upon the body:

T1 + T2 = 0

F1 ( y + (\frac{L-y}{2}) ) g sinФ - F2 x (\frac{L}{2}) x gsinФ = 0

plug in the  equations of F1 and F2 into the above equation and after simplification, we get:

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As we know that,

r = 5/9 x w

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(L^{2} - y^{2} ) . w = L^{2} . 5/9 x w

Hence,

(L^{2} - y^{2} ) = \frac{5L^{2} }{9}

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and fraction of the length of the rod submerged in water = 1 - \frac{y}{L} = 1 - 0.66 = 0.34

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