Understanding the concept:

The sound travels from the tourist to the cliff and then back to the tourist. Therefore, the total distance traveled by the sound is twice the distance between the tourist and the cliff.

Formula:

We use the formula: distance = speed × time

Given values:

• Speed of sound = v = 344 ms⁻¹
• Total time for the echo = t = 5 s

Calculations:

Let, d be the distance between the tourist and the cliff. The total distance traveled by the sound is 2d.

So,

$2d = v \cdot t \quad \Rightarrow \quad 2d = 344\ \mathrm{m\,s^{-1}} \times 5\ \mathrm{s}$

$2d = 1720\ \mathrm{m} \quad \Rightarrow \quad d = \frac{1720}{2}\ \mathrm{m} \quad \Rightarrow \quad d = 860\ \mathrm{m}$

Therefore, the distance of the cliff from the tourist is 860 meters.

Formula:

We use the wave speed formula: $v = f\lambda$, where:

• v is the speed of the wave,
• f is the frequency,
• $\lambda$ is the wavelength.

Given values:

• Speed of sound, v = 1500 ms⁻¹
• Wavelength, $\lambda$ = 45 cm = 0.45 m (since we need to use consistent units)

Rearrange the formula to solve for frequency:

$f = \frac{v}{\lambda} \quad \Rightarrow \quad f = \frac{1500\ \mathrm{m\,s^{-1}}}{0.45\ \mathrm{m}} \quad \Rightarrow \quad f = 3333.3\ \mathrm{Hz}$

Therefore, the frequency of the sound waves in seawater is approximately 3333.3 Hz.

Understanding the concept:

The sound wave travels from the surface to the seabed and then back to the surface. Therefore, the total distance traveled by the sound wave is twice the distance between the surface and the bottom of the sea.

Formula:

We use the formula: distance = speed × time

Given values:

• Speed of sound = v = 1500 ms⁻¹
• Total time for the round trip = t = 8.5 s

Calculations:

Let d be the distance between the surface and the bottom of the sea. The total distance traveled by the sound is 2d.

So,

$2d = v \cdot t \quad \Rightarrow \quad 2d = 1500\ \mathrm{m\,s^{-1}} \times 8.5\ \mathrm{s}$

$2d = 12750\ \mathrm{m} \quad \Rightarrow \quad d = \frac{12750}{2} \quad \Rightarrow \quad d = 6375\ \mathrm{m}$

Therefore, the distance from the surface to the bottom of the sea is 6375 meters.

Given Value:

Number of heartbeats in 1 minute = 69 beats

Calculations:

Frequency:

First, convert the time to seconds: 1 minute = 60 seconds

$f = \frac{\text{Number of heartbeats}}{\text{Time in seconds}} = \frac{69\ \text{beats}}{60\ \text{s}} = 1.15\ \text{Hz}$

Period:

$T = \frac{1}{f} = \frac{1}{1.15\ \text{Hz}} = 0.8696\ \text{s}$

Therefore, the frequency of the heartbeats is 1.15 Hz, and the period is 0.8696 seconds.

Formula:

We use the wave speed formula: $v = f\lambda$

Given values:

• Speed of sound = v = 343 ms⁻¹
• Minimum frequency = f_min = 20 Hz
• Maximum frequency = f_max = 20 kHz = 20,000 Hz
• Wavelength = $\lambda = ?$

Calculations:

Maximum wavelength (corresponding to minimum frequency):

$\lambda_{\max} = \frac{v}{f_{\min}} = \frac{343\ \mathrm{m\,s^{-1}}}{20\ \mathrm{Hz}} = 17.15\ \mathrm{m}$

Minimum wavelength (corresponding to maximum frequency):

$\lambda_{\min} = \frac{v}{f_{\max}} = \frac{343\ \mathrm{m\,s^{-1}}}{20{,}000\ \mathrm{Hz}} = 0.01715\ \mathrm{m}$

Therefore, the range of wavelengths for the audible frequency range is from 0.01715 m to 17.15 m.

Formula:

We use the formula: distance = speed × time

Given values:

• Speed of sound = v = 340 ms⁻¹
• Time delay = t = 3 s

Calculations:

Let d be the distance between the clouds and the ground.

So, $d = v \cdot t \quad \Rightarrow \quad d = 340\ \mathrm{m\,s^{-1}} \times 3\ \mathrm{s} \quad \Rightarrow \quad d = 1020\ \mathrm{m}$

Therefore, the distance of the clouds from the ground is 1020 meters.

Understanding the concept:

The sound travels from the SONAR device to the sea bed and then back to the device. Therefore, the total distance traveled by the sound is twice the depth of the sea.

Formula:

We use the formula: distance = speed × time

Given values:

• Speed of sound, v = 1550 ms⁻¹
• Total time for the round trip, t = 5.3 s

Calculations:

Let d be the depth of the sea. The total distance traveled by the sound is 2d.

So,

$2d = v \cdot t \quad \Rightarrow \quad 2d = 1550\ \mathrm{m\,s^{-1}} \times 5.3\ \mathrm{s}$

$2d = 8215\ \mathrm{m} \quad \Rightarrow \quad d = \frac{8215}{2}\ \mathrm{m} \quad \Rightarrow \quad d = 4107.5\ \mathrm{m}$

Therefore, the depth of the sea is 4107.5 meters.