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Grace [21]
4 years ago
13

Dolphins emit clicks of sound for communication and echolocation. A marine biologist is monitoring a dolphin swimming in seawate

r where the speed of sound is 1522 m/s. When the dolphin is swimming directly away at 7.2 m/s, the marine biologist measures the number of clicks occurring per second to be at a frequency of 2674 Hz. What is the difference (in Hz) between this frequency and the number of clicks per second actually emitted by the dolphin?
Physics
1 answer:
LenKa [72]4 years ago
6 0

Answer:

12.64968 Hz

Explanation:

v = Velocity of sound in seawater = 1522 m/s

u = Velocity of dolphin = 7.2 m/s

f' = Actual frequency = 2674 Hz

From Doppler effect we get the relation

f=f'\frac{v-u}{v}\\\Rightarrow f=2674\frac{1522-7.2}{1522}\\\Rightarrow f=2661.35032\ Hz

The frequency that will be received is 2661.35032 Hz

The difference in the frequency will be

2674-2661.35032=12.64968\ Hz

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Soloha48 [4]

Answer:

Should be B

Explanation:

As the light enters the prism at an angle, the glass causes the light to travel slower. The speed is affected by the frequency of the light. So each frequency is slowed by a different amount, causing the colors to separate.

Hope this helps!

6 0
3 years ago
A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev
tatuchka [14]

Answer:

W_f = 2.319 rad/s

Explanation:

For answer this we will use the law of the conservation of the angular momentum.

L_i = L_f

so:

I_mW_m = I_sW_f

where I_m is the moment of inertia of the merry-go-round, W_m is the initial angular velocity of the merry-go-round, I_s is the moment of inertia of the merry-go-round and the child together and W_f is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = \frac{1}{2}M_mR^2

I = \frac{1}{2}(115 kg)(2.5m)^2

I = 359.375 kg*m^2

Where M_m is the mass and R is the radio of the merry-go-round

Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2\pi rad/s

W = 3.2672 rad/s

Third, we will find the moment of inertia of both after the collision:

I_s = \frac{1}{2}M_mR^2+mR^2

I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2

I_s = 506.25kg*m^2

Finally we replace all the data:

(359.375)(3.2672) = (506.25)W_f

Solving for W_f:

W_f = 2.319 rad/s

7 0
3 years ago
Two students are watching a person riding a skateboard up and down a ramp. Each student shares what they think about the energy
avanturin [10]

Answer:

Explanation:

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By using energy conservation

KE₁ +  U₁ = KE₂ + U₂

KE₁=Kinetic energy at location 1

U₁ =Potential energy at location 1

KE₂=Kinetic energy at location 2

U₂=Potential energy at location 2

Therefore, Raymond is thinking in a right way.

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3 years ago
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kolbaska11 [484]
It decreases because it gave its momentum to the other car.
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Answer:

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