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Anna35 [415]
3 years ago
8

1. You do not need to remove the lead weights inside tires before recycling them.

Engineering
1 answer:
makvit [3.9K]3 years ago
7 0
It true -.-.-.-.-.-.-.-.-.-.-.-.-.- EZ
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A cylindrical bar of metal having a diameter of 20.9 mm and a length of 205 mm is deformed elastically in tension with a force o
natita [175]

Answer:

a. 1.91 b.  -8.13 mm

Explanation:

Modulus =stress/strain; calculating stress =F/A, hence determine the strain

Poisson's ratio =(change in diameter/diameter)/strain

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3 years ago
A rigid insulated tank is divided into 2 equal compartments by a thin rigid partition. One of the compartments contains air, ass
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there will be the answer

6 0
3 years ago
When checking for a no-star concern, you notice that an engine has no spark Technician A says to turn on the ignition engine (en
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Answer:

Technician B

Explanation:

Technician B is correct in his argument. This is because according to what he said, as the computer pulses stimuli the coil will turn on and off, promoting an increase in the voltage that will cause the fluctuation. Technician A is incorrect because the procedure he indicated imposes that the voltage is checked at the negative terminal and not at the positive.

5 0
3 years ago
Time management is a learned behavior.<br> True<br> False
larisa [96]

Answer:

true

Explanation:

3 0
3 years ago
Read 2 more answers
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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