Ask question

Ask question

All categories

12 days ago

8

A policeman in a stationary car measures the speed of approaching cars by means of an ultrasonic device that emits a sound with

a frequency of 41.2 khz. A car is approaching him at a speed of 33.0 m/s. The wave is reflected by the car and interferes with the emitted sound producing beats. What is the frequency of the beats? The speed of sound in air is 330 m/s.

2 answers:

Yuliya22 [2.4K]12 days ago

8 0

The beat frequency is nearly 9.2 kHz

$\texttt{ }$

Further explanation

Recalling the Doppler Effect formula expressed as:

$\boxed {f' = \frac{v + v_o}{v - v_s} f}$

f' = observed frequency

f = true frequency

v = velocity of sound waves

v_o = observer's speed

v_s = source's speed

Let’s proceed with the analysis!

$\texttt{ }$

Here are the specifics:

true frequency = f = 41.2 kHz

car speed = v_c = 33.0 m/s

sound speed in air = v = 330 m/s

Required:

beat frequency = Δf =?

Methodology:

First, we will determine the observed frequency via the formula of Doppler Effect as follows:

$f' = \frac{v + v_c}{v - v_c} \times f$

$f' = \frac{330 + 33}{330 - 33} \times 41.2$

$f' = \frac{363}{297} \times 41.2$

$f' = \frac{11}{9} \times 41.2$

$f' = 50 \frac{16}{45} \texttt{ kHz}$

$f' \approx 50.4 \texttt{ kHz}$

$\texttt{ }$

Then, we’ll calculate the frequency of the beats as shown:

$\Delta f = f' - f$

$\Delta f \approx 50.4 - 41.2$

$\Delta f \approx 9.2 \texttt{ kHz}$

$\texttt{ }$

Conclusion:

The frequency of the beats measures approximately 9.2 kHz

$\texttt{ }$

Learn more

Doppler Effect:
Doppler Effect Example:

$\texttt{ }$

Answer details

Grade: College

Subject: Physics

Chapter: Sound Waves

Ostrovityanka [2.2K]12 days ago

3 0

Answer:

The beats frequency measures approximately

4.4 kHz

Explanation:

The beat frequency arises from the original ultrasound frequency, $f=41.2 kHz$ , and the frequency of the sound reflected off the car, $f'$ :

$f_B = f'-f$ (1)

To calculate the frequency of the reflected sound, we apply the Doppler effect formula:

$f'=\frac{v}{v-v_s}f$

where

v = 340 m/s, the speed of sound

$v_s =33.0 m/s$ is the velocity of the car

$f=41.2 kHz$ is the frequency of the sound emitted

By substituting values,

$f'=\frac{340 m/s}{340 m/s-33.0 m/s}(41.2 kHz)=45.6 kHz$

Thus, the beat frequency (1) is

$f_B = 45.6 kHz - 41.2 kHz=4.4 kHz$

You might be interested in

The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in va

serg [2593]

Answer:

10000 V

0.00225988700565 m²

$8\times 10^{-12}\ F$

Explanation:

E = Electric field = $4\times 10^6\ V/m$

d = Distance = 2.5 mm

Q = Charge = 80 nC

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

The potential difference is calculated as

$V=Ed\\\Rightarrow V=4\times 10^6\times 2.5\times 10^{-3}\\\Rightarrow V=10000\ V$

The potential difference across the plates amounts to 10000 V

Area is determined by

$A=\dfrac{Q}{\epsilon_0E}\\\Rightarrow A=\dfrac{80\times 10^{-9}}{8.85\times 10^{-12}\times 4\times 10^6}\\\Rightarrow A=0.00225988700565\ m^2$

The area of each plate measures 0.00225988700565 m²

Capacitance is determined by

$C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.85\times 10^{-12}\times 0.00225988700565}{2.5\times 10^{-3}}\\\Rightarrow C=8\times 10^{-12}\ F$

The capacitance is $8\times 10^{-12}\ F$

4 0

1 day ago

A vector A is added to B=6i-8j. The resultant vector is in the positive x direction and has a magnitude equal to A . What is the

Yuliya22 [2420]

The answer is letter d, 8.3.

Here’s a solution for the given problem:

We have:

B = 6i - 8j

Let A be unknown; we'll denote A as = mi + nj

The resultant A+B lies along the x-axis (which implies A+B = Ki + 0j, where K is yet to be determined,

and we also know the magnitude of A+B is equivalent to the magnitude of A,

therefore, mag(A+B)=K=sqrt(m^2+n^2), or K^2 = m^2+n^2.

Using vector addition, A+B becomes (m+6)i + (n-8)j.

Since we know A+B = Ki + 0j, we can establish that:

m + 6 = K

n - 8 = 0, which gives n=8.

Thus, K^2=m^2+n^2 means (m+6)^2 = m^2 +8^2

= m^2 + 12m + 36 = m^2 + 64

which gives us 12m = 28

m = 2.33333...

Consequently, the magnitude of A is sqrt[(2.333...)^2 + 8^2] = 8.3333.

5 0

23 days ago

Read 2 more answers

A book that weighs 0.35 kilograms is kept on a shelf that’s 2.0 meters above the ground. A picture frame that weighs 0.5 kilogra

Softa [2029]

Upon calculating, I'm nearly certain that the answer is "...to a height of 1.4 meters." Happy to assist!:)

4 0

9 days ago

Read 2 more answers

13–82. the 8-kg sack slides down the smooth ramp. if it has a speed of 1.5 m> s when y = 0.2 m, determine the normal reaction

Softa [2029]

The second question necessitates a figure to provide an answer. For the initial question
The acceleration of the sack is
1.5² - 0² = 2a(0.2)
a = 5.63 m/s²
The ramp's reaction force is
F = 8 kg (5.63 m/s²)
F = 45 N
Differentiate the kinematic equation with respect to time to find the velocity's rate of increase.

4 0

9 days ago

The acceleration due to gravity on the moon is 1.6 m/s2, about a sixth that of Earth’s. Which accurately describes the weight of

kicyunya [2264]

Response:

The question is not fully provided; here is the complete context:

Gravity's acceleration on the moon is 1.6 m/s², roughly one-sixth that of Earth's. What is the accurate description of an object's weight on the moon?

A. An object on the moon is lighter by a factor of 1/6 compared to Earth.

B. An object on the moon is heavier by a factor of 1/6 compared to Earth.

C. An object on the moon is six times lighter than on Earth.

D. An object on the moon is six times heavier than on Earth.

The correct choice is:

An object on the moon is six times lighter than on Earth. (C)

Explanation:

The acceleration resulting from gravity indicates how a gravitational force impacts an object, causing it to accelerate. This is a vectorial quantity because it possesses both magnitude and direction, measured in the unit of m/s². On Earth, this gravitational acceleration is represented by the letter g and its value is approximately 9.8m/s².

The larger size of the Earth in comparison to the moon causes its gravitational acceleration to be about six times greater than that of the moon, resulting in the moon's gravitational acceleration being approximately 1.6m/s².

Next, weight refers to the product of mass and gravity's acceleration. This reflects the gravitational pull acting upon a mass, which is also measured in Newtons, similar to force.

Weight = m × g (N)

From the weight formula, we can see that weight corresponds directly to mass and gravitational acceleration:

weight ∝ mass;

weight ∝ gravitational acceleration.

This implies that if gravitational acceleration increases, weight increases as well, and vice versa.

For instance, let's calculate the weights of a 10kg object on both Earth and the moon.

Gravitational acceleration on Earth (g₁) = 9.8m/s².

Gravitational acceleration on the moon (g₂) = 1.6m/s².

On Earth:

weight = m × g₁ = 10 × 9.8 = 98 N.

On the moon:

weight = m × g₂ = 10 × 1.6 = 16 N.

From the above example, since the acceleration due to gravity on the moon is 1/6 that of Earth, the weight of a 10kg object on the moon is approximately six times lighter (16 N) than its weight on Earth (98 N).

5 0

9 days ago

Read 2 more answers