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denpristay [2]
2 years ago
13

Two automobiles are equipped with the same singlefrequency horn. When one is at rest and the other is moving toward the first at

20 m/s , the driver at rest hears a beat frequency of 9.0 Hz.
Requried:
What is the frequency the horns emit?
Physics
1 answer:
Kamila [148]2 years ago
8 0

Answer: f ≈ 8.5Hz

Explanation: The phenomenon known as <u>Doppler</u> <u>Shift</u> is characterized as a change in frequency when one observer is stationary and the source emitting the frequency is moving or when both observer and source are moving.

For a source moving and a stationary observer, to determine the frequency:

f_{0} = f_{s}.\frac{c}{c-v_{s}}

where:

f_{0} is frequency of observer;

f_{s} is frequency of source;

c is the constant speed of sound c = 340m/s;

v_{s} is velocity of source;

Rearraging for frequency of source:

f_{0} = f_{s}.\frac{c}{c-v_{s}}

f_{s} = f_{0}.\frac{c-v_{s}}{c}

Replacing and calculating:

f_{s} = 9.(\frac{340-20}{340})

f_{s} = 9.(0.9412)

f_{s} = 8.5

<u>Frequency the horns emit is 8.5Hz.</u>

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7 0
2 years ago
All ball is thrown up with a vertical velocity of 54 m/s and a horizontal velocity of 39 m/s. Calculate how many seconds it will
VikaD [51]

5.5 s

Explanation:

The time it takes for the ball to reach its maximum height can be calculated using

v_y = v_{0y} - gt \Rightarrow t = \dfrac{v_{0y}}{g}

since v_y = 0 at the top of its trajectory. Plugging in the numbers,

t = \dfrac{(54\:\text{m/s})}{(9.8\:\text{m/s}^2)} = 5.5\:\text{s}

8 0
3 years ago
Waves are observed passing under a dock. Wave crests are 8.0 meters apart. The time for a complete wave to pass by is 4.0 second
ki77a [65]
To answer that question, we don't care what the highest and lowest
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often those pass by us.

You said it takes 4 seconds for a complete wave to pass by.
Through the sheer power of intellect, I'm able to take that information
and calculate that  1/4  of the wave passes by in 1 second.

There's your frequency . . .  1/4 per second, or  0.25 Hz.
6 0
2 years ago
What is the contour interval of this map? a. 20 feet b. -20 feet c. 60 feet 11​
timofeeve [1]

Answer:

c. 60 feet is the correct answer

Explanation:

what is the contour interval of this map? a.20 b.-20 c. 60 feet 11

8 0
3 years ago
A uniformly charged conducting sphere of 1.1 m diameter has a surface charge density of 6.2 µC/m2. (a) Find the net charge on th
ira [324]

Answer:

(a) q = 2.357 x 10⁻⁵ C

(b) Φ = 2.66 x 10⁶ N.m²/C

Explanation:

Given;

diameter of the sphere, d = 1.1 m

radius of the sphere, r = 1.1 / 2 = 0.55 m

surface charge density, σ = 6.2 µC/m²

(a)  Net charge on the sphere

q = 4πr²σ

where;

4πr² is surface area of the sphere

q is the net charge on the sphere

σ is the surface charge density

q = 4π(0.55)²(6.2 x 10⁻⁶)

q = 2.357 x 10⁻⁵ C

(b) the total electric flux leaving the surface of the sphere

Φ = q / ε

where;

Φ is the total electric flux leaving the surface of the sphere

ε is the permittivity of free space

Φ = (2.357 x 10⁻⁵) / (8.85 x 10⁻¹²)

Φ = 2.66 x 10⁶ N.m²/C

8 0
2 years ago
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