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3 years ago

14

Technician A says that the paper test could detect a burned valve. Technician B says that a grayish white stain on the engine co

uld be a coolant leak. Who is right

1 answer:

Bumek [7]3 years ago

6 0

Answer:

Both Technician A and B are correct.

Explanation: Both are correct, but keep note that different color coolants leave different color stains.

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The scale of the blueprint tells us the<br> of drawing to real space?

klasskru [66]

Answer:

yes

Explanation:

blueprint of the construction is a prediction of project its is slightly auto cad

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3 years ago

12-32 The Clear Brook High School band is holding a car wash as a fund-raiser to buy new equipment. The average time to wash a c

goldenfox [79]

Answer:

69

Explanation:

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4 years ago

0-0 what is Boyle’s Law

Dima020 [189]

A law stating that the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature.

Hope it helps

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3 years ago

Atmospheric air at 25 °C and 8 m/s flows over both surfaces of an isothermal (179C) flat plate that is 2.75m long. Determine the

vekshin1

Answer:

Re=100,000⇒Q=275.25 $\frac{W}{m^2}$

Re=500,000⇒Q=1,757.77 $\frac{W}{m^2}$

Re=1,000,000⇒Q=3060.36 $\frac{W}{m^2}$

Explanation:

Given:

For air $T_∞$ =25°C ,V=8 m/s

For surface $T_s$ =179°C

L=2.75 m ,b=3 m

We know that for flat plate

$Re$ ⇒Laminar flow

$Re>30\times10^5$ ⇒Turbulent flow

<u> Take Re=100,000:</u>

So this is case of laminar flow

$Nu=0.664Re^{\frac{1}{2}}Pr^{\frac{1}{3}}$

From standard air property table at 25°C

Pr= is 0.71 ,K=26.24 $\times 10^{-3}$

So $Nu=0.664\times 100,000^{\frac{1}{2}}\times 0.71^{\frac{1}{3}}$

Nu=187.32 ( $\dfrac{hL}{K_{air}}$ )

187.32= $\dfrac{h\times2.75}{26.24\times 10^{-3}}$

⇒h=1.78 $\frac{W}{m^2-K}$

heat transfer rate =h $(T_∞-T_s)$

=275.25 $\frac{W}{m^2}$

<u> Take Re=500,000:</u>

So this is case of turbulent flow

$Nu=0.037Re^{\frac{4}{5}}Pr^{\frac{1}{3}}$

$Nu=0.037\times 500,000^{\frac{4}{5}}\times 0.71^{\frac{1}{3}}$

Nu=1196.18 ⇒h=11.14 $\frac{W}{m^2-K}$

heat transfer rate =h $(T_∞-T_s)$

=11.14(179-25)

= 1,757.77 $\frac{W}{m^2}$

<u> Take Re=1,000,000:</u>

So this is case of turbulent flow

$Nu=0.037Re^{\frac{4}{5}}Pr^{\frac{1}{3}}$

$Nu=0.037\times 1,000,000^{\frac{4}{5}}\times 0.71^{\frac{1}{3}}$

Nu=2082.6 ⇒h=19.87 $\frac{W}{m^2-K}$

heat transfer rate =h $(T_∞-T_s)$

=19.87(179-25)

= 3060.36 $\frac{W}{m^2}$

7 0

3 years ago

A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm (0.60 in. × 0.75 in.) is pulled in tension with 44,500

lukranit [14]

Answer:

The resulting strain is $1.39\times 10^{-3}$ .

Explanation:

A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm

Force, F = 44,500 N

Th elastic modulus of Cu to be 110 GPa

The resulting strain is given by the formula as follows :

$\epsilon=\dfrac{F}{AE}$

E is elastic modulus of Cu is are of cross section

$\epsilon=\dfrac{44500}{15.2\times 19.1\times 10^{-6}\times 110\times 10^9}\\\\\epsilon=1.39\times 10^{-3}$

So, the resulting strain is $1.39\times 10^{-3}$ .

6 0

4 years ago