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melamori03 [73]
2 years ago
7

4. A light string is attatched to a heavy rope, and the whole thing is pulled tight. A wave is sent along the light string. When

it hits the heavy rope, compared to the wave on the string, the wave that propagates along the rope has the same (A) frequency (B) wavelength (C) both frequency and wavelength (D) neither?
Physics
1 answer:
ale4655 [162]2 years ago
7 0

Answer:

The correct answer to the question is (A)

When it hits the heavy rope, compared to the wave on the string, the wave that propagates along the rope has the same (A) frequency

Explanation:

The speed of a wave in a string is dependent on the square root of the tension ad inversely proportional  to the square root of the linear density of the string. Generally, the speed of a wave through a spring is dependent on the elastic and inertia properties of the string

v = \sqrt{ \frac{T}{\mu } } =  \sqrt{ \frac{T}{m/L } }

Therefore if the linear density of the heavy rope is four times that of light rope the velocity is halved and since

v = f×λ therefore  v/2 = f×λ/2

Therefore the wavelength is halved, however the frequency remains the same as continuity requires the frequency of the incident pulse vibration to be transmitted to the denser medium for the wave to continue as the wave is due to vibrating particles from a source for example

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If we decrease the amount of force applied to an object, and all other factors remain the same, the amount of work completed wil
Nat2105 [25]
A ) decrease.
B ) increase.
C ) increase, then decrease.
D ) not change.

The answer is A) decrease

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3 years ago
at certain times the demand for electric energy is low and electric energy is used to pump water to a reservoir 45 m above the g
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The mass of water that must be raised is 5.25\cdot 10^7 kg

Explanation:

Since the process is 70% efficiency, the power in output to the turbine can be written as

P_{out} = 0.70 P_{in}

where P_{in} is the power in input.

The power in input can be written as

P_{in} = \frac{W}{t}

where

W is the work done in lifting the water

t = 3 h = 10,800 s is the time elapsed

The work done in lifting the water is given by

W=mgh

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m is the mass of water

g=9.8 m/s^2 is the acceleration of gravity

h = 45 m is the height at which the water is lifted

Combining the three equations together, we get:

P_{out} = 0.70 \frac{mgh}{t}

Where

P_{out} = 150 MW = 150\cdot 10^6 W

And solving for m, we find:

m=\frac{Pt}{0.70gh}=\frac{(1.50\cdot 10^6)(10800)}{(0.70)(9.8)(45)}=5.25\cdot 10^7 kg

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3 0
3 years ago
A refrigerator is used to cool water from 23°C to 5°C in a continuous manner. The heat rejected in the condenser is 570 kJ/min a
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COP=2.58

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density  = 1 kg/L

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Power in put W= 2.65 KW

From first law of thermodynamics

U = W+ q

q = Heat absorbed

U = internal energy

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Q=0.091 L/min

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