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2 years ago

13

which of the following tools is used for measuring small diameter holes which a telescoping gauge cannot fit into? A. telescopin

g gauges B. Feeler Guages C. Dial indicator D. Small hold gauges

1 answer:

Alex2 years ago

7 0

Answer:

C. Dial indicator

Explanation:

This meassers small diameters

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The town of Mustang, TX is concerned that waste heat discharged from a new up- stream power plant will decimate the minnow popul

vitfil [10]

Answer:

Yes the water will be safe at the point of cooling water discharge

Explanation:

Power losses in plant= 350- 350×0.35=227.5MW

Rate of heat rejection to stream= 0.75× 227.5= 170.625MW

Rate of heat rejection= rate of flow of water× c × ΔT

170625000= 150000000× 4.186 × (Final temperature- 20)

Final temperature= 20.3 ◦C

The final temperature of stream will be 20.3 ◦C. Thechange is very small so the minnows will be able to handle this temperature.

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2 years ago

A turbojet aircraft flies with a velocity of 800 ft/s at an altitude where the air is at 10 psia and 20 F. The compressor has a

nika2105 [10]

Answer:

Pressure = 115.6 psia

Explanation:

Given:

v=800ft/s

Air temperature = 10 psia

Air pressure = 20F

Compression pressure ratio = 8

temperature at turbine inlet = 2200F

Conversion:

1 Btu =775.5 ft lbf, $g_{c}$ = 32.2 lbm.ft/lbf.s², 1Btu/lbm=25037ft²/s²

Air standard assumptions:

$c_{p}$ = 0.0240Btu/lbm.°R, R = 53.34ft.lbf/lbm.°R = 1717.5ft²/s².°R 0.0686Btu/lbm.°R

k= 1.4

Energy balance:

$h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} + \frac{v_{a} ^{2} }{2}\\$

As enthalpy exerts more influence than the kinetic energy inside the engine, kinetic energy of the fluid inside the engine is negligible

hence $v_{a} ^{2} = 0$

$h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} \\h_{1} -h_{a} = - \frac{v_{1} ^{2} }{2} \\ c_{p} (T_{1} -T_{a})= - \frac{v_{1} ^{2} }{2} \\(T_{1} -T_{a}) = - \frac{v_{1} ^{2} }{2c_{p} }\\ T_{a}=T_{1} + \frac{v_{1} ^{2} }{2c_{p} }$

$T_{1}$ = 20+460 = 480°R

$T_{a} =480+ \frac{(800)(800}{2(0.240)(25037}$ = 533.25°R

Pressure at the inlet of compressor at isentropic condition

$P_{a } =P_{1}(\frac{T_{a} }{T_{1} }) ^{k/(k-1)}$

$P_{a}$ = $(10)(\frac{533.25}{480}) ^{1.4/(1.4-1)}$ = 14.45 psia

$P_{2}= 8P_{a} = 8(14.45) = 115.6 psia$

4 0

2 years ago

Read 2 more answers

A civil engineer is studying a left-turn lane that is long enough to hold seven cars. Let X be the number of cars in the line at

BartSMP [9]

Answer:

a) C= 1/120

b) P(X>=5) = 0.333

Explanation:

The attached file contains the explanation for the answers

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2 years ago

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Which one of the following activities is not an example of incident coordination

Lady bird [3.3K]

Directing, ordering, or controlling

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2 years ago

determine the position d of the 6- kn load so that the average normal stress in each rod is the same.

Zinaida [17]

The load is placed at distance 0.4 L from the end of $$12 \mathrm{~mm}^{2}$ area.

<h3>What is meant by torque?</h3>

The force that can cause an object to rotate along an axis is measured as torque. Similar to how force accelerates an item in linear kinematics, torque accelerates an object in an angular direction. A vector quantity is torque.

Let the beam is of length L

Now the stress on both the end is the same now we can say that torque on the beam due to two forces must be zero

$N_1 * x=N_2 *(L-x)$

also, we know that stress at both ends are same

$\frac{N_1}{12}=\frac{N_2}{8}$

$2 * N_1=3 * N_2$

Now from two equations we have

$$\frac{3}{2} N_2 * x=N_2 *(L-x)$

solving the above equation we have

$$x=\frac{2}{5} L$

so the load is placed at distance 0.4 L from the end of $$12 \mathrm{~mm}^{2}$ area.

The complete question is:

47. the beam is supported by two rods ab and cd that have cross-sectional areas of $$$12mm^2$ and $$$8mm^2$ , respectively. determine the position d of the 6-kn load so that the average normal stress in each rod is the same.

To learn more about torque refer to:

brainly.com/question/20691242

#SPJ4

7 0

1 year ago