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dedylja [7]
3 years ago
13

____ is the rate of change in velocity.

Physics
2 answers:
Dimas [21]3 years ago
6 0

<u>Answer:</u>

<u>Acceleration</u> is the rate of change in velocity.

<u>Explanation:</u>

<u>Velocity,v- </u>

"It is the change in a body's displacement or change in speed,d over the time,t."

  • v=s/sec,
  • <u>Units:</u> meter/second.

<u>Acceleration,a-</u>

"When the body has a varying velocity,v across a given time frame,t is called as the acceleration,a."

  • a=v/t,
  • <u>Unit:</u> meter/second².
astra-53 [7]3 years ago
4 0
The answer is Acceleration.
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\frac{\sigma_{cv}}{m}= (s_2-s_1) \geq  0 \ \ \ -------> \ \ \ 1

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If the value is less than zero; then we conclude that the assumption is wrong.

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s_2-s_1 = s^0(T_2) - s^0(T_1)-R \ In ( \frac{P_2}{P-1}) \ \ \  ------->  \ \ \ 2

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s_1, s_2 , s^0(T_2), s^0(T1) are specific entropies

R = universal gas constant

P_1 = pressure at location 1

P_2 = pressure at location 2

We obtain the specific properties of air at temperature at T_1 = (67°C + 273)K = 340 K from the table A-22 ( Ideal gas properties of air)

s^0(T1) = 1.8279 kJ/kg.K

We also obtain the specific properties of air at temperature T_2 = 22°C + 273) K = 295 K

From the table A- 22

s^0(T_2) = 1.68515 kJ/kg . K

R = \frac{8.314 kJ}{28.97 kg.K}

P_1 = 0.95 bar

P_2 = 0.8 bar

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s_2-s_1 = s^0(T_2) -s^0(T_1)-R \ In (\frac{P_2}{P_1} )

s_2-s_1 = 1.68515 -1.8279-\frac{8.314}{28.97}  \ In (\frac{0.8}{0.95} )

s_2-s_1 = 1.68515 -1.8279+ 0.0493

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Equating our result to equation (1)

s_2-s_1 \geq 0\\-0.0934 \leq 0

Therefore , our assumption is wrong and the direction of flow is said to be from right to left.

We therefore conclude that the direction of flow is from right to left.

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