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ExtremeBDS [4]
3 years ago
11

An eagle flying at 35 m/s emits a cry whose frequency is 440 Hz. A blackbird is moving in the same direction as the eagle at 10

m/s. (Assume the speed of sound is 343 m/s.)
(a) What frequency does the blackbird hear (in Hz) as the eagle approaches the blackbird?
Hz
(b) What frequency does the blackbird hear (in Hz) after the eagle passes the blackbird?
Hz
Physics
1 answer:
Akimi4 [234]3 years ago
7 0

Answer:

a)  F=475.7Hz

b)  F'=410.899Hz

Explanation:

From the question we are told that:

Velocity of eagle V_1=35m/s

Frequency of eagle F_1=440Hz

Velocity of Black bird V_2=10m/s

Speed of sound s=343m/s

a)

Generally the equation for Frequency is mathematically given by

 F=f_0(\frac{v-v_2}{v-v_1})

 F=440(\frac{343-10}{343-35})

 F=475.7Hz

b)

Generally the equation for Frequency is mathematically given by

 F'=f_0(\frac{v+v_2}{v+v_1})

 F'=440(\frac{343+10}{343+35})

 F'=410.899Hz

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Leni [432]

Answer: The differences between terrestrial planets and the giant planets are s follows-

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4 years ago
Our eyes are typically 6 cm apart. Suppose you are somewhat unique, and yours are 7.50 cm apart. You see an object jump from sid
Strike441 [17]

Answer:

Distance of the object from eye is approx 4.52 m

Explanation:

As we know that the object subtend a small angle on both the eyes which is given as

\theta = 0.995 degree

now we know that the distance between two eyes is given as

d = 7.50 cm

so we have

angle = \frac{arc}{Radius}

so here the radius is same as the distance from eye while arc is the distance between two eyes

so we have

0.95 \frac{\pi}{180} = \frac{7.50}{R}

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3 years ago
Water falls without splashing at a rate of 0.200 L/s from a height of 3.60 m into a 0.730 kg bucket on a scale. If the bucket is
dimaraw [331]

Answer:

15.106 N

Explanation:

From the given information,

The weight of the bucket can be calculated as:

W_b = m_bg =  \\ \\  W_b = (0.730 \  kg) ( 9.80 \ m/s^2) \\ \\ W_b = 7.154 \ N

The mass of the water accumulated in the bucket after 3.20s is:

m_w= (0.20 \ L/s) ( 3.20)s

m _w=0.64 \ kg

To determine the weight of the water accumulated in the bucket, we have:

W_w = m_w g

W_w = ( 0.64  \ kg )(9.80\  m  \  /s^2)

W_w = 6.272 \ N

For the speed of the water before hitting the bucket; we have:

v = \sqrt{2gh}

v = \sqrt{2*9.80 \ m/s^2 * 3.60 \ m}

v = 8.4 m/s

Now, the force required to stop the water later when it already hit the bucket is:

F = v ( \dfrac {dm}{dt} )

F = (8.4 \ m/s)( 0.200 \ L/s)

F = 1.68 N

Finally, the reading scale is:

F_{scale = 7.154 N + 6.272 N + 1.68 N

= 15.106 N

6 0
3 years ago
Read 2 more answers
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