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Andru [333]
2 years ago
6

Submarine a travels horizontally at 11.0 m/s through ocean water. it emits a sonar signal of frequency f 5 5.27 3 103 hz in the

forward direction. submarine b is in front of submarine a and traveling at 3.00 m/s relative to the water in the same direction as submarine
a. a crewman in submarine b uses his equipment to detect the sound waves ("pings") from submarine
a. we wish to determine what is heard by the crewman in submarine
b. (a) an observer on which submarine detects a frequency f 9 as described by equation 17.19? (b) in equation 17.19, should the sign of vs be positive or negative? (c) in equation 17.19, should the sign of vo be positive or negative? (d) in equation 17.19, what speed of sound should be used? (e) find the frequency of the sound detected by the crewman on submarine b
Physics
1 answer:
xeze [42]2 years ago
6 0
Velocity of submarine A is vs = 11.0m/s
frequency emitted by submarine A. F = 55.273 × 10∧3HZ
Velocity of submarine B = vO = 3.00m/s
The given equation is
f' = ((V + vO) ((v - vS)) × f
The observer on submarine detects the frequency f'.
The sign of vO should be positive as the observer of submarine B is moving away from the source of submarine A.
The speed of the sound used in seawater is 1533m/s
The frequency which is detected by submarine B is 
fo = fs (V -vO/ v +vs)
= 53.273 × 10∧3hz) ((1533 m/s - 4.5 m/s)/ (1533 m/s +11 m/s)
fo = 5408 HZ
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Answer:

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Explanation:

The detailed step by step solution to this problems can be found in the attachment below. The solution for part (a) was divided into two: the motion of the body from point A to point B and from point B to point C. The total energy in the system is gotten from the initial gravitational potential energy. This energy becomes transformed into the work done against friction and the work done in compression the spring. A work of 398J was done in overcoming friction over a distance of 6.00m. The energy used in doing so is lost as friction is not a conservative force. This leaves only 43J of energy which compresses the spring. On expansion the spring does a work of 43J back on the block is only enough to push it over a distance of 0.65m stopping short of 5.35m from point B.

Thank you for reading and I hope this is helpful to you.

4 0
2 years ago
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Answer:

The newton’s second law is F=ma

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Explanation:

Given that,

The equivalent formulas for velocity, acceleration, and force using the relationships covered for UCM, Newton’s Laws, and Gravitation.

We know that,

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The velocity is equal to the rate of position of the object.

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The acceleration is equal to the rate of velocity of the object.

a=\dfrac{dv}{dt}....(II)

Newton’s second Laws

The force is equal to the change in momentum.

In mathematically,

F=\dfrac{d(p)}{dt}

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The force is equal to the product of mass of objects and divided by square of distance.

In mathematically,

F=\dfrac{Gm_{1}m_{2}}{r^2}

Where, m₁₂ = mass of first object

m= mass of second object

r = distance between both objects

Hence, The newton’s second law is F=ma

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2 years ago
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