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3 years ago

What is the measurement of this square? A.075 B. 1.75 C. 0.34

Engineering

1 answer:

Katena32 [7]3 years ago

7 0

Answer:

0.75 in

Explanation:

The 1 inch measure has 16 divisions.

Find the measure of 1 division as: 1/16 = 0.0625 in

The side length of the square has 12 divisions

1 division = 0.0625 in

12 divisions = 0.0625*12 = 0.75 in

The correct answer choice is A. 0.75.

Answer choices B is incorrect because from the diagram, it is clear that the side length of this square is less that 1 inch.

Answer choice C is incorrect because 0.34 in will mean 0.34/0.0625 divisions. This is 5 divisions, yet the square side length covers 12 divisions.

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Answer:

a. true

Explanation:

Firstly, we need to understand what takes places during the compression process in a quasi-equilibrium process. A quasi-equilibrium process is a process in during which the system remains very close to a state of equilibrium at all times. When a compression process is quasi-equilibrium, the work done during the compression is returned to the surroundings during expansion, no exchange of heat, and then the system and the surroundings return to their initial states. Thus a reversible process.

While for a non-quasi equilibrium process, it takes more work to move the piston against this high-pressure region.

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3 years ago

What is the primary function of NCEES?

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3 years ago

Consider a half-wave rectifier circuit with a triangular-wave input of LaTeX: 6V6 V peak-to-peak amplitude and zero average valu

Varvara68 [4.7K]

The question is not complete. We are supposed to find the average value of v_o.

Answer:

v_o,avg = 0.441V

Explanation:

Let t1 and t2 be the start and stop times of the output waveforms. Thus, from the diagram i attached, using similar triangles, we have;

3/(T/4) = 0.7/t1

So, 12/T = 0.7/t1

So, t1 = 0.7T/12

t1 = 0.0583 T

Also, from symmetry of triangles,

t2 = T/2 - t1

So, t2 = T/2 - 0.0583 T

t2 = 0.4417T

Average of voltage output is;

v_o,avg = (1/T) x Area under small triangle

v_o,avg = (1/T) x (3 - 0.7) x (T/4 - t1)

v_o,avg = (1/T) x (2.3) x (T/4 - 0.0583 T)

v_o,avg = (1/T) x 2.3 x 0.1917T

T will cancel out to give;

v_o,avg = 0.441V

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2 years ago