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Cloud [144]
2 years ago
10

Question 5(Multiple Choice Worth 2 points)

Engineering
1 answer:
Liono4ka [1.6K]2 years ago
6 0

Answer:

I believe it is sensor time

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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.

3 0
3 years ago
A negative pressure respirator brings fresh air to you through a hose<br>A) True<br>B)False​
madreJ [45]

Answer:FALSE

Explanation: A negative pressure respirator is a respiratory system which is known to have a low air pressure inside the mask when compared to the air pressure on the outside during Inhalation.

Most of the personal protective equipment (PPE) which are in use in various industries are examples of Negative pressure respirator device,any leak or damage done to the device will allow the inflows of harmful and toxic Air into the person's respiratory system. AIR SUPPLY SYSTEMS ARE KNOWN TO SUPPLY FRESH UNCONTAMINATED AIR THROUGH AIR STORED INSIDE COMPRESSED CYLINDERS OR OTHER SOURCES AVAILABLE.

8 0
3 years ago
Read 2 more answers
What type of oil pressure gauge should be used when
liq [111]

Answer:

The mechanical gauge would be the one for the job

Explanation:

6 0
2 years ago
Tion A. Classwork
pantera1 [17]

Answer:

ambot...

Explanation:

hahahahahhahahahahaha

5 0
3 years ago
Problem Statement: Air flows at a rate of 0.1 kg/s through a device as shown below. The pressure and temperature of the air at l
Tema [17]

Answer:

The answer is "+9.05 kw"

Explanation:

In the given question some information is missing which can be given in the following attachment.

The solution to this question can be defined as follows:

let assume that flow is from 1 to 2 then

Q= 1kw

m=0.1 kg/s

From the steady flow energy equation is:

m\{n_1+ \frac{v^2_1}{z}+ gz_1 \}+Q= m \{h_2+ \frac{v^2_2}{2}+ gz_2\}+w\\\\\ change \ energy\\\\0.1[1.005 \times 800]-1= 0.01[1.005\times 700]+w\\\\w= +9.05 \ kw\\\\

If the sign of the work performed is positive, it means the work is done on the surrounding so, that the expected direction of the flow is right.

8 0
2 years ago
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