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Lilit [14]
3 years ago
9

Two submarines are underwater and approaching each other head-on.Sub A has a speed of 4 m/s and sub B has aspeed of 14 m/s. Sub

A sends out a1910 Hz sonar wave that travels at a speedof 1522 m/s.(a) What is the frequency of the sound detected by subB (to the nearest Hz)?Hz(b) Part of the sonar wave is reflected from B and returns to A.What frequency does A detect for this reflected wave (to thenearest Hz)?Hz
Physics
1 answer:
Triss [41]3 years ago
7 0

Answer:

1933 Hz

1956 Hz

Explanation:

v_0 = Velocity of object

v_s = Velocity of source

v = Velocity of sound

From Doppler effect

f=f'\dfrac{v+v_0}{v-v_s}\\\Rightarrow f=1910\dfrac{1522+14}{1522-4}\\\Rightarrow f=1932.64822134\ Hz=1933\ Hz

The frequency of the sound detected by subB is 1933 Hz

f=f'\dfrac{v+v_0}{v-v_s}\\\Rightarrow f=1932.64822134\dfrac{1522+4}{1522-14}\\\Rightarrow f=1955.7169\ Hz=1956\ Hz

The frequency of the sound detected by subA is 1956 Hz

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A 0.5 kg basketball moving 5 m/s to the right collides with a 0.05 kg tennis
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Answer:

A. 1.4 m/s to the left

Explanation:

To solve this problem we must use the principle of conservation of momentum. Let's define the velocity signs according to the direction, if the velocity is to the right, a positive sign will be introduced into the equation, if the velocity is to the left, a negative sign will be introduced into the equation. Two moments will be analyzed in this equation. The moment before the collision and the moment after the collision. The moment before the collision is taken to the left of the equation and the moment after the collision to the right, so we have:

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where:

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where:

m1 = mass of the basketball = 0.5 [kg]

v1 = velocity of the basketball before the collision = 5 [m/s]

m2 = mass of the tennis ball = 0.05 [kg]

v2 = velocity of the tennis ball before the collision = - 30 [m/s]

v3 =  velocity of the basketball after the collision [m/s]

v4 = velocity of the tennis ball after the collision = 34 [m/s]

Now replacing and solving:

(0.5*5) - (0.05*30) = (0.5*v3) + (0.05*34)

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3 years ago
A cart of mass m = 0.12 kg moves with a speed v = 0.45 m/s on a frictionless air track and collides with an identical cart that
lina2011 [118]

Answer:

0.006075Joules

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The final kinetic energy of the system is expressed as;

KE = 1/2(m1+m2)v²

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get the final velocity using the law of conservation of momentum

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KE = 1/2(0.12+0.12)(0.225)²

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KE = 0.12*0.050625

KE = 0.006075Joules

Hence the final kinetic energy of the system is 0.006075Joules

5 0
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