$$
c = 343, mathrm{m/s}
$$

Now we have to convert it into $mathrm{km/s}$:

$$
c = dfrac{343, mathrm{m/s}}{1000} = 0.343, mathrm{km/s} = dfrac{1}{2.92}, mathrm{km/s}
$$

Since $2.92mathrm{s} approx 3, mathrm{s}$ we can conclude that sound travels around $1, mathrm{km}$ in $3, mathrm{s}$

Now we have to see what happens in the case of miles and we can conclude that it travels around $1, mathrm{m}$ in $mathrm{5, mathrm{S}}$ which later lead us to the answer that the number of seconds shall be divided by five.

$textbf{(b) }$ Wavelength will increase because speed is increasing and frequency is constant

$textbf{(b) }$ Wavelength will decrease

$textbf{(c) }$ Wave velocity remains unchanged

$textbf{(c) }$ Amplitude remains unchanged

$$
lambda = 4l
$$

Wavelength is also given by:

$$
lambda = dfrac{v}{f}
$$

$$
rightarrow 4l = dfrac{v}{f}
$$

$v$ from the previous equation will be:

$$
v = 4fl
$$

When speed increase, temperature will increase as well that is why we have determined it, and what we can conclude from the last equation is that when $l$ is unchained and $v$ increases $f$ increases as well since they are directly proportional, therefore, the pitch will increase as well.

Since beat frequency is increasing, the frequencies are getting farther apart

$t_{1} = 3, mathrm{s}$

$v = 343, mathrm{m/s}$ which is speed of sound in air

$dfrac{l}{2}=?$

Speed is given by:

$$
v = dfrac{l}{t}
$$

$l$ from the previous equation will be:

$$
l = vt
$$

When we put known values into the previous equation we get:

$$
l = 343, mathrm{m/s} cdot 3, mathrm{s}
$$

Required distance will then be:

$$
dfrac{l}{2} = dfrac{343, mathrm{m/s} cdot 3, mathrm{s}}{2}
$$

$$
boxed{dfrac{l}{2} approx 2.6cdot 10^{2}, mathrm{m}}
$$

Therefore $lambda = 1.1$ m

$$
{color{#c34632}v = lambdacdot f} = 1.1cdot 4700 = 5170
$$

$f = 261.6, mathrm{Hz}$

$T = 25, mathrm{^oC}$

$1493, mathrm{m/s}$ – which is speed of sound in water

$lambda = ?$

Wavelength is given by:

$$
lambda = dfrac{v}{f}
$$

When we put known values into the previous equation we get:

$$
lambda = dfrac{1493, mathrm{m/s}}{261.6, mathrm{Hz}}
$$

$$
boxed{lambda = 5.707, mathrm{m}}
$$

$150, mathrm{dB}$ sound level that he experiences

$textit{a.}$, Since rock concert is around $110, mathrm{dB}$, required reduction will be:

$$
150, mathrm{dB} – 110, mathrm{dB} = boxed{40, mathrm{dB}}
$$

$textit{b.}$, Barely audible whisper is $10, mathrm{dB}$, therefore actual level will be:

$$
40, mathrm{dB} + 10, mathrm{dB} = boxed{50, mathrm{dB}}
$$

$textit{b.}$, $50, mathrm{dB}$

$80, mathrm{dB}$ sound level of rock band playing

$textit{a.}$, $100, mathrm{dB}$ sound level of another rock band:

As we have said previously, when there is $20, mathrm{dB}$ of sound level will increase pressure by $10$ factor, therefore, since the increasing is from $80, mathrm{dB}$ to $100, mathrm{dB}$ pressure will be $boxed{$10$}$ times greater.

$textit{b.}$, In this case, sound level is $120, mathrm{dB}$
Now the pressure will increase from $100, mathrm{dB}$ to $120, mathrm{dB}$, therefore it will be $boxed{$100$}$ times greater.

$textit{b.}$, Pressure will be $100$ times greater.

$f = 4, mathrm{Hz = dfrac{1}{s}}$

$lambda = 0.5, mathrm{m}$

$v = ?$

Speed is given by:

$$
v = lambda f
$$

When we put known values into the previous equation we get:

$$
v = 0.5, mathrm{m} cdot 4, mathrm{dfrac{1}{s}}
$$

$$
boxed{v = 2, mathrm{m/s}}
$$

Speed of sound increases by 0.6 m/s for every 1$text{textdegree}$ C increase in temperature

We know that the speed of sound at $20text{textdegree}$ C is 343 m/s

Therefore speed of sound at $30text{textdegree}$ C will be $343+0.6cdot 10 = 343+6 = 349$ m/s

Duration = $dfrac{text{Distance}}{text{Speed}} = dfrac{152}{349}approx0.436$ seconds

We know that the speed of sound at $20text{textdegree}$ C is 343 m/s

Therefore speed of sound at $15text{textdegree}$ C will be $343-0.6cdot 5 = 343-3 = 340$ m/s

Distance = $text{Duration}times{text{Speed}} =1times340 = 340$ m

$f = 4.25, mathrm{MHz} = 4.25 cdot 10^{6}, mathrm{Hz}$

$v = 1.50, mathrm{km/s} = 1.50 cdot 10^{3}, mathrm{m/s}$

$lambda = ?$

Wavelength is given by:

$$
lambda = dfrac{v}{f}
$$

When we put known values into the previous equation we get:

$$
f = dfrac{1.5 cdot 10^{3}, mathrm{m/s}}{4.25 cdot 10^{6}, mathrm{Hz}}
$$

$$
boxed{lambda = 0.353, mathrm{m}}
$$

$t_{1} = 1.74, mathrm{s}$

$t_{2} = 2.36, mathrm{s}$

$T = 25, mathrm{^oC}$

$v = 1875, mathrm{m/s}$

$v_{s} = 1533, mathrm{m/s}$ – speed of sound in seawater

$textit{a.}$, $d=?$

Speed is given by depth per time:

$$
v_{s} = dfrac{d}{t}
$$

Therefore, depth will be:

$$
d = v_{s} cdot t
$$

Since $t$ is time for trip in one way it will be:

$$
begin{align*}
t &= dfrac{t_{1}}{2}\
&= dfrac{1.74, mathrm{s}}{2}\
&= 0.87, mathrm{s}\
end{align*}
$$

Therefore, the depth will be:

$$
d = 1533, mathrm{m/s} cdot 0.87, mathrm{s}
$$

$$
boxed{d = 1300, mathrm{m}}
$$

Speed is given by:

$$
v = dfrac{l}{t}
$$

Therefore $l$ will be:

$$
; = vt
$$

$t$ is one-way time in the mud, but in order to determine it, we must determine round-trip first:

$$
t_{round-trip} = 2.36, mathrm{s} – 1.74, mathrm{s} = 0.62, mathrm{s}
$$

$t$ will then be:

$$
begin{align*}
t &= dfrac{t_{round-trip}}{2}\
&= dfrac{0.62, mathrm{s}}{2}\
&= 0.31, mathrm{s}\
end{align*}
$$

Finally, thickness of the mud will be:

$$
l = 1875, mathrm{m/s} cdot 0.31, mathrm{s}
$$