3 years ago

Since the passing of the Utah GDL laws in 1999:

Engineering

1 answer:

wlad13 [49]3 years ago

7 0

The answer is b, I hope this helps you

You might be interested in

Multiple Choice

Ymorist [56]

Answer:

Sealing agent

Explanation:

Generally, when we have water leaks in almost any building or equipment, we use a sealant. However, this sealant could be of different types depending on the peculiarity of the leakage.

Thus, the correct answer is sealing agent.

5 0

2 years ago

Read 2 more answers

Question # 3

tino4ka555 [31]

Answer:

I think it's False.

Apologies if I am wrong.

4 0

3 years ago

Read 2 more answers

The gas expanding in the combustion space of a reciprocating engine has an initial pressure of 5 MPa and an initial temperature

Anit [1.1K]

Answer:

a). Work transfer = 527.2 kJ

b). Heat Transfer = 197.7 kJ

Explanation:

Given:

$P_{1}$ = 5 Mpa

$T_{1}$ = 1623°C

= 1896 K

$V_{1}$ = 0.05 $m^{3}$

Also given $\frac{V_{2}}{V_{1}} = 20$

Therefore, $V_{2}$ = 1 $m^{3}$

R = 0.27 kJ / kg-K

$C_{V}$ = 0.8 kJ / kg-K

Also given : $P_{1}V_{1}^{1.25}=C$

Therefore, $P_{1}V_{1}^{1.25}$ = $P_{2}V_{2}^{1.25}$

$5\times 0.05^{1.25}=P_{2}\times 1^{1.25}$

$P_{2}$ = 0.1182 MPa

a). Work transfer, δW = $\frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}$

$\left [\frac{5\times 0.05-0.1182\times 1}{1.25-1} \right ]\times 10^{6}$

= 527200 J

= 527.200 kJ

b). From 1st law of thermodynamics,

Heat transfer, δQ = ΔU+δW

= $\frac{mR(T_{2}-T_{1})}{\gamma -1}+ \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}$

= $\left [ \frac{\gamma -n}{\gamma -1} \right ]\times \delta W$

= $\left [ \frac{1.4 -1.25}{1.4 -1} \right ]\times 527.200$

= 197.7 kJ

6 0

3 years ago

A 4140 steel shaft, heat-treated to a minimum yield strength of 100 ksi, has a diameter of 1 7/16 in. The shaft rotates at 600 r

velikii [3]

Answer:

Explanation:

4140-40 I’d pick wood

I hope this helps! :)

4 0

3 years ago

Read 2 more answers

Explain when it is appropriate to use Tier II?

garik1379 [7]

????? what do u mean tier 2

3 0

2 years ago