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fgiga [73]
2 years ago
14

Technician A says that mechanical shifting controls can wear out over time. Technician B says that vacuum control rubber diaphra

gms can deteriorate over time. Who is correct
Engineering
1 answer:
diamong [38]2 years ago
8 0

Based on the information, both technician A and technician B are correct.

<h3>How to depict the information?</h3>

From the information given, Technician A says that mechanical shifting controls can wear out over time.

Technician B says that vacuum control rubber diaphragms can deteriorate over time.

In this case, both technicians are correct as the information depicted is true.

Learn more about technicians on:

brainly.com/question/1548867

#SPJ12

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Natural Gas Corporation obtains a federal license to operate a gas pipeline through a certain area of Oregon. The Oregon state l
pentagon [3]

Answer:

b. the supremacy clause

Explanation:

This is clearly a violation of the supremacy clause of the constitution. The supremacy clause makes the constitution and federal laws made under the constitutional authority the supreme law of the united state. And in a case where there is a conflicting state law, as we have here with Oregon, the federal law is supposed to take priority.

So in this case where the federal law and the state law do not agree, the feral law has the power to override the law of the state. So oregon has violated the supremacy clause

6 0
2 years ago
What is the next measurement after 2' -6" on the architect's scale?
Diano4ka-milaya [45]

Answer: I am not for sure

Explanation:

6 0
3 years ago
A wire of diameter d is stretched along the centerline of a pipe of diameter D. For a given pressure drop per unit length of pip
JulsSmile [24]

Answer:

Part A: (d/D=0.1)

DeltaV percent=42.6%

Part B:(d/D=0.01)

DeltaV percent=21.7%

Explanation:

We are going to use the following volume flow rate equation:

DeltaV=\frac{\pi * DeltaP}{8*u*l}(R^{4}-r^{4} -\frac{(R^{2}-r^{2})}{ln\frac{R}{r}}^{2})

Above equation can be written as:

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(\frac{r}{R} )^{4}+\frac{(1-(\frac{r}{R} )^{2})}{ln\frac{r}{R}}^{2})

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(\frac{d}{D} )^{4}+\frac{(1-(\frac{d}{D})^{2})}{ln\frac{d}{D}}^{2})

First Consider no wire i.e d/D=0

Above expression will become:

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0)^{4}+\frac{(1-(0)^{2})}{ln0}^{2})

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}

Part A: (d/D=0.1)

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0.1)^{4}+\frac{(1-(0.1)^{2})}{ln0.1}^{2})

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}*0.574

DeltaV percent=\frac{(\frac{\pi*R^{4}*DeltaP}{8*u*l})-\frac{\pi *R^{4}*DeltaP}{8*u*l}*0.574}{\frac{\pi*R^{4}*DeltaP}{8*u*l} }*100

DeltaV percent=\frac{1-0.574}{1}*100

DeltaV percent=42.6%

Part B:(d/D=0.01)

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0.01)^{4}+\frac{(1-(0.01 )^{2})}{ln0.01}^{2})

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}*0.783

DeltaV percent=\frac{(\frac{\pi *R^{4}*DeltaP}{8*u*l})-\frac{\pi *R^{4}*DeltaP}{8*u*l}*0.783}{\frac{\pi *R^{4}*DeltaP}{8*u*l} }*100

DeltaV percent=\frac{1-0.783}{1}*100

DeltaV percent=21.7%

5 0
3 years ago
What is the single largest contributor to increasing takeoff or landing distance?
hodyreva [135]

An uphill slope improves the take-off ground run, and a downhill slope increases the landing ground run.

<h3>What is uphill slope?</h3>

Driving uphill suggests climbing a “positive” six percent slope . Driving downhill, the “rise” exists as a  drop, so there is a “negative,” or downhill, slope (Figure B). When dealing with slope, a positive slope simply indicates uphill and a negative slope indicates downhill. An uphill slope improves the take-off ground run, and a downhill slope increases the landing ground run.

If something or someone lives moving downhill or is downhill, they exist moving down a slope or are located toward the bottom of a hill. He headed downhill toward the river. adverb. If you communicate that something exists going downhill, you mean that it is becoming worse or less prosperous.

To learn more about uphill slope refer to:

brainly.com/question/13361896

#SPJ4

3 0
2 years ago
A 20 dBm power source is connected to the input of a directional coupler having a coupling factor of 20 dB, a directivity of 35
lukranit [14]

Answer:

P_O = 0.989 watt = 19.9 dBm

Explanation:

Given data:

P_1 power = 20 dBm  = 0.1 watt

coupling factor is 20dB

Directivity = 35 dB

We know that

coupling factor = 10 log \frac{P_1}{P_f}

solving for  final power

20 = 10 log\frac{P_1}{P_f}

2 = log \frac{P_1}{P_f}

100 = \frac{0.1}{P_f}

P_f = 0.001 watt = 0 dBm

Directivity D =  10 \frac{P_f}{P_b}

35 = 10 \frac{0.001}{P_b}

P_b = 3.162 \times 10^{-7} wattt

output Power  = P_1 -P_f - P_b

                       = 0.1 - 0.001 - 3.162 \times 10^{-7}

P_O = 0.989 watt = 19.9 dBm

6 0
3 years ago
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