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Grace [21]
3 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]3 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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A rock is dropped from a height of 3,245 m. If we ignore air resistance, how fast will it be travelling after it falls for 3.4 s
Marta_Voda [28]

Answer:

<u>954.4m/s</u>

Explanation:

For a free falling object,it has constant acceleration and a changing velocity.

By using the velocity-time formula, the velocity can be obtained.

The height the rock travelled is the distance.

From,

Velocity (v) = Distance (d) / Time(t)

v = 3245m/3.4s

v = <u>954.4m/s</u>

That js the answer I got. Hope it's right.

3 0
2 years ago
The electron gun in a television tube is used to accelerate electrons with mass 9.109 × 10−31 kg from rest to 3 × 107 m/s within
zaharov [31]

Answer:

Electric field, E = 40608.75 N/C

Explanation:

It is given that,

Mass of electrons, m=9.1\times 10^{-31}\ kg

Initial speed of electron, u = 0

Final speed of electrons, v=3\times 10^7\ m/s

Distance traveled, s = 6.3 cm = 0.063 m

Firstly, we will find the acceleration of the electron using third equation of motion as :

a=\dfrac{v^2-u^2}{2s}

a=\dfrac{(3\times 10^7)^2}{2\times 0.063}

a=7.14\times 10^{15}\ m/s^2

Now we will find the electric field required in the tube as :

ma=qE

E=\dfrac{ma}{q}

E=\dfrac{9.1\times 10^{-31}\times 7.14\times 10^{15}}{1.6\times 10^{-19}}

E = 40608.75 N/C

So, the electric field required in the tube is 40608.75 N/C. Hence, this is the required solution.

3 0
2 years ago
You throw a rock upward. The rock is moving upward, but it is slowing down. If we define the ground as the origin, the position
lyudmila [28]

Answer:

A. Positive, positive

Explanation:

As we defined the ground is the origin, as long as the rock is above the ground, its position would always be positive.

As for velocity, as long as the rock is moving upwards, its direction is also upward, so its velocity is always positive. The velocity will become negative when the rock falls down.

6 0
3 years ago
What is the basic formula for the harmonic of a standing wave
Pie

fn=nf1 is the formula

4 0
3 years ago
Read 2 more answers
Someone just help me
Zanzabum
C according to my calculations
3 0
2 years ago
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