A. B. D. C. D, A, A, C, B, B, D, D
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Answer:
71.4583 Hz
67.9064 N
Explanation:
L = Length of tube = 1.2 m
l = Length of wire = 0.35 m
m = Mass of wire = 9.5 g
v = Speed of sound in air = 343 m/s
The fundamental frequency of the tube (closed at one end) is given by

The fundamental frequency of the wire and tube is equal so he fundamental frequency of the wire is 71.4583 Hz
The linear density of the wire is

The fundamental frequency of the wire is given by

The tension in the wire is 67.9064 N
By applying Newton's second law of motion;
ma = mg - T
Where,
m = mass; a = downward accelerations (+ve value) or upward acceleration (-ve value); g = gravitational acceleration; T = tension.
For the current case, the velocity is constant therefore,
a = 0
Then,
0 = mg - T
T = mg = 115*9.81 = 1128.15 N
Tension in the cable is 1128.15 N.
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3. electric compressor: a device that <span>acts as a pump, circulating refrigerant throughout the refrigerator
4. the inside of the fridge and the food becomes colder
5. the coolants becomes a hot, high-pressure gas
6. as coolant transfers thermal energy to the air outside, it turns back into a liquid
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