A stone is dropped from rest into a well. The sound of the splash is heard exactly 2.00s later. Find the depth of the well if the air temperature is 10.0°c ?

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A stone is dropped from rest into a well. The sound of the splash is heard exactly 2.00s later. Find the depth of the well if the air temperature is 10.0°c ?

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Answer 1 · 2015-08-26T13:21:26

$674$ $674$ meters

Explanation:

Although there will be some variation depending upon air humidity (for example) the speed of sound relative to temperature (measured in degrees Celsius) can be determined by the empirical formula:
$XXX s = 331 \frac{meters}{sec.} + 0.6 \frac{meters}{sec.} \cdot c$ $color(white)("XXX")s=331" meters"/"sec."+0.6 "meters"/"sec."*c$
where $c$ $c$ is the temperature in degrees Celsius.

For a speed of $s$ $s$ and time of $t$ $t$ ,
the distance $d$ $d$ is
$\cdot X X X d = s \times t$ $color(white)(*XXX")d = sxxt$

Therefore we have
$XXX D = 331 + 0.6 (10) \frac{meters}{second} \times 2 seconds$ $color(white)("XXX")D =331+0.6(10) "meters"/"second" xx 2 "seconds"$

$XXXX = 674 meters$ $color(white)("XXXX")= 674 "meters"$

Answer 2 · 2015-08-27T02:05:16

$19.8 m$ $19.8"m"$

Explanation:

For the 1st part the stone is falling under gravity. Let $t_{1}$ $t_1$ be the time from release to splashdown.

$d$ $d$ =depth

So $d = \frac{1}{2} g t_{1}^{2}$ $d=(1)/(2)"g"t_(1)^2$ $" "$ $(1)$ $color(red)((1))$

After splashdown the sound travels back up. Using $330 m/s$ $330"m/s"$ for the speed of sound this gives:

$d = 330 \times t_{2}$ $d=330xxt_2$ $" "$ $(2)$ $color(red)((2))$

We also know that:

$t_{1} + t_{2} = 2 s$ $t_1+t_2=2"s"$ $(3)$ $" "color(red)((3))$

Combining $(1) and (2)$ $color(red)((1))"and"color(red)((2))$ we get:

$330 t_{2} = \frac{1}{2} g . t_{1}^{2}$ $330t_2=1/2g.t_1^2$

From $(3) \Rightarrow$ $color(red)((3))rArr$

$t_{2} = (2 - t_{1})$ $t_2=(2-t_1)$

So:

$330 (2 - t_{1}) = \frac{1}{2} . g t_{1}^{2}$ $330(2-t_1)=1/2."g"t_1^2$

$660 - 330 t_{1} = \frac{1}{2} . g t_{1}^{2}$ $660-330t_1=1/2."g"t_1^2$

$\frac{1}{2} g t_{1}^{2} + 330 t_{1} - 660 = 0$ $1/2"g"t_1^2+330t_1-660=0$

Using the quadratic formula to solve for $t_{1}$ $t_1$ :

$t_{1} = \frac{- 330 \pm \sqrt{{(330)}^{2} - 4 \times \frac{9.8}{2} \times (- 660)}}{9.8}$ $t_1=(-330+-sqrt((330)^2-4xx9.8/2xx(-660)))/(9.8)$

Ignoring the -ve root this gives:

$t_{1} = 1.94 s$ $t_1=1.94"s"$

So $t_{2} = 2 - 1.94 = 0.06 s$ $t_2=2-1.94=0.06"s"$

So $d = 330 \times t_{2} = 330 \times 0.06 = 19.8 m$ $d=330xxt_2=330xx0.06=19.8"m"$

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A stone is dropped from rest into a well. The sound of the splash is heard exactly 2.00s later. Find the depth of the well if the air temperature is 10.0°c ?

2 Answers

Explanation:

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