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Anna [14]
2 years ago
8

Technician A says that a fully charged battery is less likely to freeze than a discharged battery. Technician B says that the st

ate of charge has no impact on freezing. Who is correct
Engineering
1 answer:
zheka24 [161]2 years ago
8 0
Answer : Technician B is correct
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For welding the most important reason to use jigs and fixtures in a welding shop is to
diamong [38]

Answer:

Reduce manufacturing costs.

Explanation:

Hope This Helps

Have A Great Day

7 0
2 years ago
What is the governing ratio for thin walled cylinders?
Ann [662]

Answer:

The governing ratio for thin walled cylinders is 10 if you use the radius. So if you divide the cylinder´s radius by its thickness and your result is more than 10, then you can use the thin walled cylinder stress formulas, in other words:

  • if \frac{radius}{thickness} >10 then you have a thin walled cylinder

or using the diameter:

  • if \frac{diameter}{thickness} >20 then you have a thin walled cylinder
3 0
3 years ago
Consider an infinitely thin flat plate of chord c at an angle of attack α in a supersonic flow. The pressure on the upper and lo
amm1812

Answer:

X_cp = c/2

Explanation:

We are given;

Chord = c

Angle of attack = α

p u (s) = c 1

​p1(s)=c2,

and c2 > c1

First of all, we need to find the resultant normal force on the plate and the total moment about leading edge.

I've attached the solution

4 0
3 years ago
What tool is used to measure aggregates in laying tools<br>​
denpristay [2]

Answer:

Sieve Shaker

Explanation:

6 0
2 years ago
Air is compressed in a reversible, isothermal, steady- flow process from 15 psia, 100°F to 100 psia. Calculate the work of compr
mixas84 [53]

Answer:

|W|=169.28 KJ/kg

ΔS = -0.544 KJ/Kg.K

Explanation:

Given that

T= 100°F

We know that

1 °F = 255.92 K

100°F = 310 .92 K

P _1= 15 psia

P _1= 100 psia

We know that work for isothermal process  

W=mRT\ln \dfrac{P_1}{P_2}

Lets take mass is 1 kg.

So work per unit mass

W=RT\ln \dfrac{P_1}{P_2}

We know that for air R=0.287KJ/kg.K

W=RT\ln \dfrac{P_1}{P_2}

W=0.287\times 310.92\ln \dfrac{15}{100}

W= - 169.28 KJ/kg

Negative sign indicates compression

|W|=169.28 KJ/kg

We know that change in entropy at constant volume

\Delta S=-R\ln \dfrac{P_2}{P_1}

\Delta S=-0.287\ln \dfrac{100}{15}

ΔS = -0.544 KJ/Kg.K

3 0
3 years ago
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