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saul85 [17]
2 years ago
13

15

Engineering
1 answer:
AveGali [126]2 years ago
4 0

Answer:

b

Explanation:

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Answer: The accelerator and the brakes.

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Does anybody know what plane this is? i saw it the other day doing a low pass through my community
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it is what it is

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3 years ago
Find the percent change in cutting speed required to give an 80% reduction in tool life when the value of n is 0.12.
vaieri [72.5K]

Answer:21.3%

Explanation:

Given

80 % reduction in tool life

According to Taylor's tool life

VT^n=c

where V is cutting velocity

T=tool life of tool

80 % tool life reduction i.e. New tool Life is 0.2T

Thus

VT^{0.12}=V'\left ( 0.2T\right )^{0.12}

V'=\frac{V}{0.2^{0.12}}

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6 0
2 years ago
Air is compressed by a 40-kW compressor from P1 to P2. The air temperature is maintained constant at 25°C during this process a
AlexFokin [52]

Answer:

the rate of entropy change of the air is -0.1342 kW/K

the assumptions made in solving this problem

- Air is an ideal gas.

- the process is isothermal ( internally reversible process ). the change in internal energy is 0.

- It is a steady flow process

- Potential and Kinetic energy changes are negligible.

Explanation:

Given the data in the question;

From the first law of thermodynamics;

dQ = dU + dW ------ let this be equation 1

where dQ is the heat transfer, dU is internal energy and dW is the work done.

from the question, the process is isothermal ( internally reversible process )

Thus, the change in internal energy is 0

dU = 0

given that; Air is compressed by a 40-kW compressor from P1 to P2

since it is compressed, dW = -40 kW

we substitute into equation 1

dQ = 0 + ( -40 kW )

dQ = -40 kW

Now, change in entropy of air is;

ΔS_{air = dQ / T

given that T = 25 °C = ( 25 + 273.15 ) K = 298.15 K

so we substitute

ΔS_{air =  -40 kW / 298.15 K

ΔS_{air =  -0.13416 ≈ -0.1342 kW/K

Therefore, the rate of entropy change of the air is -0.1342 kW/K

the assumptions made in solving this problem

- Air is an ideal gas.

- the process is isothermal ( internally reversible process ). the change in internal energy is 0.

- It is a steady flow process

- Potential and Kinetic energy changes are negligible.

7 0
2 years ago
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