ENGINEERING COLLEGE

An equal-tangent sag vertical curve is designed for 45 mi/h. The low point is 237 ft from the PVC at station 112 37 and the final offset at the PVT is 19.355 ft. If the PVC is at station 110 00, what is the elevation difference between the PVT and a point on the curve at station 111 00

Answers

Answer 1
Answer:

Answer:

18.722 ft

Explanation:

The elevation difference between the PVT and a point on the curve at station

111 + 00

attached below is a detailed solution to the problem

Δelevation ( elevation difference )

= Yt - Y

= 19.355 - 0.632 = 18.722 ft

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Answers

Answer:

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Explanation:

Please see attachment for step by step guide

Which of the following expressions causes an implicit conversion between types? Assume variable x is an integer, t is a float, and name is a string.Group of answer choices7.5 + (x / 2)x + 2 * x"Hello, " + str(name)print(str(t))

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brainly.com/question/31063278

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Karen just checked her bank balance and the balance was 1111.11. She is kind of freaking out and saved the receipt, thinking that it must mean something. Karen is probably just suffering from

Answers

Answer:

chronic stoner syndrome

Explanation:

"the universe just sends us messages sometimes mannnn, you just have to be ready to listen to them" lol

In a refrigerator, heat is transferred from a lower-temperature medium (the refrigerated space) to a higher-temperature one (the kitchen air). Is this a violation of the second law of thermodynamics

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What is the function of deaerator in thermal power plant? ​

Answers

Answer:

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Explanation:

Air expands adiabatically through a nozzle from a negligible initial velocity to a final velocity of 300 m/s, what is the temperature drop of the air, if air is assumed to be an ideal gas for which CP = (7/2)R?

Answers

The temperature drop of air if air is assumed to be an ideal gas for which C_p = ⁷/₂R is; Δt = 1546 K

We are given;

Final velocity; v₂ = 300 m/s

C_p = ⁷/₂R

At constant pressure, the change in enthalpy is;

Δh = C_p × Δt

Now, from first law of thermodynamics;

h₂ + (v₂²/2) = h₁ + (v₁²/2)

We are told initial velocity is negligible and as such v₁ = 0 m/s

Thus;

h₂ + (v₂²/2) = h₁ + 0

(h₁ - h₂) =  (v₂²/2)

Thus; Δh = v₂²/2

Finally;

C_p × Δt = v₂²/2

Δt = v₂²/2/(C_p)

Δt =  (300²/2)/(⁷/₂R)

where R is ideal gas constant = 8.314 Kj/kg.mol

Thus;

Δt = (300²/2)/(⁷/₂ × 8.314)

Δt = 1546 K

Read more at; brainly.com/question/24188841

Answer:

ΔH+U²/2=0

and

ΔH=C{p×ΔT

∴to get the temperature drop of air, you make ΔT subject of the formula

ΔT=-U²/2Cp

    =-300²/2×(7)/(2)×8.314

∴ΔT=-1546K

Explanation:
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