A sonar signal of frequency 1.00 106 Hz has a wavelength of 1.40 mm in water.

(a) What is the speed of the signal in water?
(b) What is its period in water?
(c) What is its period in air?

period=1/frequency in air, and water

speed=wavelength*frequency

(a) Well, I'm no fish, but I do know that the speed of a sonar signal in water can be calculated using the formula v = λf, where v is the speed, λ is the wavelength, and f is the frequency. In this case, the frequency is 1.00 * 10^6 Hz and the wavelength is 1.40 mm. So, let's crunch the numbers: v = (1.40 mm) * (1.00 * 10^6 Hz) = 1.40 * 10^3 m/s. Voila!

(b) Now, to find the period in water, we can use the formula T = 1/f, where T is the period and f is the frequency. In this case, the frequency is still 1.00 * 10^6 Hz, so the period would be T = 1 / (1.00 * 10^6 Hz) = 1.00 * 10^-6 s.

(c) Ah, now to calculate the period in air. Since the speed of sound in air is different from that in water, the period will be different too. Using the same formula T = 1/f and the given frequency of 1.00 * 10^6 Hz, the period in air would be T = 1 / (1.00 * 10^6 Hz) = 1.00 * 10^-6 s. Looks like the period stays the same regardless of the medium!

And there you have it, the speed in water is 1.40 * 10^3 m/s, the period in water and air is 1.00 * 10^-6 s.

To calculate the speed, period, and frequency of the sonar signal, we can use the formula:

Speed = Frequency x Wavelength

(a) To find the speed of the signal in water:
Given: Frequency = 1.00 x 10^6 Hz, Wavelength = 1.40 mm

Converting the wavelength to meters: 1.40 mm = 1.40 x 10^-3 m

Using the formula, we have:
Speed = Frequency x Wavelength
Speed = 1.00 x 10^6 Hz x 1.40 x 10^-3 m

Calculating:
Speed = 1.40 x 10^3 m/s

Therefore, the speed of the signal in water is 1.40 x 10^3 m/s.

(b) To find the period of the signal in water:
The period is the time taken for one complete cycle of the wave.

Using the formula:
Period = 1 / Frequency

Given Frequency = 1.00 x 10^6 Hz

Calculating:
Period = 1 / (1.00 x 10^6 Hz)

Therefore, the period of the signal in water is 1.00 x 10^-6 s.

(c) To find the period of the signal in air:
The frequency remains the same in air, but the speed changes as the wave travels through different mediums.

Using the formula:
Period = 1 / Frequency

Given Frequency = 1.00 x 10^6 Hz

Calculating:
Period = 1 / (1.00 x 10^6 Hz)

Therefore, the period of the signal in air is also 1.00 x 10^-6 s.

To solve these questions, we can use the wave equation:

v = λ * f

where v is the velocity (speed), λ is the wavelength, and f is the frequency of the signal.

(a) To find the speed of the signal in water:

We are given the frequency f = 1.00 * 10^6 Hz and the wavelength λ = 1.40 mm.

Converting the wavelength to meters:
λ = 1.40 mm = 1.40 * 10^-3 m

Now we can use the wave equation:
v = λ * f
v = (1.40 * 10^-3 m) * (1.00 * 10^6 Hz)
v ≈ 1400 m/s

So, the speed of the signal in water is approximately 1400 m/s.

(b) To find the period of the signal in water:

The period (T) of a wave is the time it takes for one complete cycle.

We know that frequency (f) is the inverse of the period (T):
f = 1 / T

Rearranging the equation:
T = 1 / f

Given f = 1.00 * 10^6 Hz, we can calculate the period:
T = 1 / (1.00 * 10^6 Hz)
T = 1.00 * 10^(-6) s (or 1 μs)

So, the period of the signal in water is 1 μs.

(c) To find the period of the signal in air:

The speed of sound in air is approximately 343 m/s.

Using the wave equation, we can find the period:
v = λ * f
T = λ / v

Given the wavelength in water (λ = 1.40 mm = 1.40 * 10^-3 m) and the speed of sound in air (v = 343 m/s), we can calculate the period:
T = (1.40 * 10^-3 m) / 343 m/s
T ≈ 4.08 * 10^-6 s (or 4.08 μs)

So, the period of the signal in air is approximately 4.08 μs.