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Hatshy [7]
3 years ago
10

Two submarines are underwater and approaching each other head-on. Sub A has a speed of 15.8 m/s and sub B has a speed of 8.63 m/

s. Sub A sends out a 1570-Hz sonar wave that travels at a speed of 1522 m/s.
(a) What is the frequency detected by sub B?
(b) Part of the sonar wave is reflected from B and returns to A.
What frequency does A detect for this reflected wave?
Physics
1 answer:
Evgen [1.6K]3 years ago
8 0

Answer:

a) The frequency detected by the submarine B, f' = 1595 Hz

b) The frequency of the reflected wave detected by the A is, f'' = 6021 Hz

Explanation:

Given data,

The speed of submarine A, v = 15.8 m/s

The speed of submarine B, v' = 8.63 m/s

The SONAR wave travels at a speed of V = 1522 m/s

The formula for Doppler effect is,

                              f' = \frac{V+v'}{V-v} f

Substituting the given values above,

                              f' = \frac{1522+8.63}{1522-15.8} 1570

                             f' = 1595 Hz

The reflected SONAR has the frequency of, f' = 1595 Hz

Therefore, the frequency of the reflected wave at A is given by the formula,

                             f'' = \frac{V+v}{V-v'} f'

                             f'' = \frac{1522+15.8}{1522-8.63} 1595

                             f'' = 6021 Hz

Hence, the reflected wave has the frequency of f'' = 6021 Hz

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<h3>What are compressed gases?</h3>

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Since the cylinder valve is open and the gas is collected at atmospheric pressure, less gas will be collected because some of the gases will escape.

Since, the carbon dioxide not liquefy under pressure compared to butane, the cylinders used to store carbon dioxide will have thicker walls than those of butane.

Learn more about compressed gases at: brainly.com/question/518065

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2 years ago
A 5.0-kilogram sphere, starting from rest, falls freely 22 meters in 3.0 seconds near the surface of a planet. Compared to the a
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Answer:

C) one-half as great

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In this case, the sphere starts from rest, so v_0=0. Replacing the given values and solving for g':

g'=\frac{2\Delta x}{t^2}\\g'=\frac{2(22m)}{(3s)^2}\\g'=4.89\frac{m}{s^2}

The acceleration due to gravity near Earth's surface is g=9.8\frac{m}{s^2}. So, the acceleration due to gravity near the surface of the planet is approximately one-half of the acceleration due to gravity near Earth's surface.

5 0
3 years ago
An empty graduated cylinder weighs 55.26 g. When filled with 48.1 mL of an unknown liquid, it weighs 92.39 g. The density of the
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<h2>Density of the unknown liquid is 771.93 kg/m³</h2>

Explanation:

An empty graduated cylinder weighs 55.26 g

Weight of empty cylinder = 55.26 g = 0.05526 kg

Volume of liquid filled = 48.1 mL = 48.1 x 10⁻⁶ m³

Weight of cylinder plus liquid = 92.39 g = 0.09239 kg

Weight of liquid = 0.09239 - 0.05526

Weight of liquid = 0.03713 kg

We have

        Mass = Volume x Density

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A particle of mass 4.5 × 10-8 kg and charge +5.4 μC is traveling due east. It enters perpendicularly a magnetic field whose magn
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Answer:

0.00970 s

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The centripetal force that causes the charge to move in a circular motion = The force exerted on the charge due to magnetic field

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θ = angle between the velocity of the charge and the magnetic field = 90°, sin 90° = 1

F = qvB

Centripetal force responsible for circular motion = mv²/r = mvw

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The centripetal force that causes the charge to move in a circular motion = The force exerted on the charge due to magnetic field

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mw = qB

w = (qB/m) = (5.4 × 10⁻⁶ × 2.7)/(4.5 × 10⁻⁸)

w = 3.24 × 10² rad/s

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Time = (angular displacement)/w

Angular displacement = π rads (half of a circle; 2π/2)

Time = (π/324) = 0.00970 s

Hope this Helps!!!

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