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Ratling [72]
3 years ago
12

You drop a stone down a well that is 19.60 m deep. How long is it before you hear the splash? The speed of sound in air is 343 m

/s and air resistance is negligible.
Physics
1 answer:
ki77a [65]3 years ago
3 0

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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aliya0001 [1]

Answer:

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β= 21.8° , countercockwise from the positive x-axis: Direction of the second displacement

Explanation:

We find the x-y components for the given vectors:

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j:unit vector in y direction

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D₂: Displacement Vector 2

R= resulta displacement vector

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We propose the vector equation for sum of vectors:

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-51.99i+142.83j+D₂x(i)-D₂y(j) = 103.23i+80.65j

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D₂x=103.23+51.99=155.22 cm

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D₂y=142.83-80.65=62.18 cm

Magnitude and direction of the second displacement

D_{2} =\sqrt{(D_{x})^{2} +(D_{y} )^{2}  }

D_{2} =\sqrt{(155.22)^{2} +(62.18 )^{2}  }

D₂= 167.21 cm

Direction of the second displacement

\beta = tan^{-1} \frac{D_{y}}{D_{x} }

\beta = tan^{-1} \frac{62.18}{155.22 }

β= 21.8°

D₂= 167,21 cm : Magnitude  of the second displacement

β= 21.8.° , countercockwise from the positive x-axis: Direction of the second displacement

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Answer:

The SI units of A, B and C are :

m,\ m/s\ and\ m/s^2                  

Explanation:

The position x, in meters, of an object is given by the equation:

x=A+Bt+Ct^2

Where

t is time in seconds

We know that the unit of x is meters, such that the units of A, Bt and Ct^2 must be meters. So,

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b=\dfrac{m}{s}=m/s

  • Ct^2=m

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So, the SI units of A, B and C are :

m,\ m/s\ and\ m/s^2

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Answer:

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v = Final velocity

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