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2 years ago

Troy must keep track of the amount of refrigerant he uses from a 50-pound cylinder to ensure that accurate

Engineering

1 answer:

IgorLugansk [536]2 years ago

7 0

Answer:

Amount of gas still in cylinder = 28 pound

Explanation:

Given:

Amount of gas in cylinder = 50 pound

Amount of gas used in Ms. Jones system = 13 pound

Amount of gas used in client system = 9 pound

Find:

Amount of gas still in cylinder

Computation:

Amount of gas still in cylinder = Amount of gas in cylinder - Amount of gas used in Ms. Jones system - Amount of gas used in client system

Amount of gas still in cylinder = 50 - 13 - 9

Amount of gas still in cylinder = 28 pound

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OA bloom is smaller than a bar

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Explanation:

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2 years ago

Quinn’s relatives relayed a story about putting on a headset and seeing a digital world that they could walk around in and explo

Kryger [21]

Answer:

I know it is C)Virtual reality

Explanation:

Look at the clues

story about putting on a headset ( virtual reality head set!)

seeing a digital world (A virtual reality world)

they could walk around in (Fake walking you are basically jogging in place)

explore in order to see what ancient Benin looked like (Looking at a real place only digitally)

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2 years ago

Read 2 more answers

Compute the solution to x + 2x + 2x = 0 for Xo = 0 mm, vo = 1 mm/s and write down the closed-form expression for the response.

Nutka1998 [239]

Answer:

β = $\frac{c}{\sqrt{km} }$ = 0.7071 ≈ 1 ( damping condition )

closed-form expression for the response is attached below

Explanation:

Given : x + 2x + 2x = 0 for Xo = 0 mm and Vo = 1 mm/s

computing a solution :

M = 1,

c = 2,

k = 2,

Wn = $\sqrt{\frac{k}{m} }$ = $\sqrt{2}$

next we determine the damping condition using the damping formula

β = $\frac{c}{\sqrt{km} }$ = 0.7071 ≈ 1

from the condition above it can be said that the damping condition indicates underdamping

attached below is the closed form expression for the response

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2 years ago

An equal-tangent sag vertical curve (with a negative initial and a positive final grade) is designed for 55 mi/h. The PVI is at

Varvara68 [4.7K]

Answer:

The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

Explanation:

Length of curve is given as

$L=2(PVT-PVI)\\L=2(242+30-240+00)\\L=2(230)\\L=460 ft$

$G_2$ is given as

$G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%$

The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as

$A=\frac{L}{K}\\A=\frac{460}{115}\\A=4$

A is given as

$-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%$

With initial grade, the elevation of PVC is

$E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\$

The station is given as

$St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\$

Low point is given as

$x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft$

The station of low point is given as

$St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\$

The elevation is given as

$E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft$

So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

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3 years ago