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AlexFokin [52]
3 years ago
11

Select the correct answer.

Engineering
2 answers:
andreev551 [17]3 years ago
8 0

Answer:

c

Explanation:

WARRIOR [948]3 years ago
3 0

Answer:

Explanation:

b

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A compound machine contains three simple machines with IMAs of 2, 4 and 5, respectively. What is the overall ideal mechanical ad
anygoal [31]

Answer:

Overall ideal mechanical advantage of the machine = 40

Explanation:

Given:

Ideal mechanical advantage of three machine = 2, 4, 5

Find:

Overall ideal mechanical advantage of the machine

Computation:

Overall ideal mechanical advantage of the machine = 2 × 4× 5

Overall ideal mechanical advantage of the machine = 40

3 0
3 years ago
A complete mix of an activated sludge system without primary clarification is used for treatment of municipal wastewater with a
Hitman42 [59]

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Explanation:

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2 years ago
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tensa zangetsu [6.8K]

Answer:

HUH?

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A heat engine operates between 2 reservoirs at TH and 18oC. The heat engine receives 17,000 kJ/h from the high temperature reser
lisabon 2012 [21]

Answer:

See explaination

Explanation:

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4 0
3 years ago
Two common methods of improving fuel efficiency of a vehicle are to reduce the drag coefficient and the frontal area of the vehi
qaws [65]

Answer:

\Delta V = 209.151\,L, \Delta C = 217.517\,USD

Explanation:

The drag force is equal to:

F_{D} = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot A

Where C_{D} is the drag coefficient and A is the frontal area, respectively. The work loss due to drag forces is:

W = F_{D}\cdot \Delta s

The reduction on amount of fuel is associated with the reduction in work loss:

\Delta W = (F_{D,1} - F_{D,2})\cdot \Delta s

Where F_{D,1} and F_{D,2} are the original and the reduced frontal areas, respectively.

\Delta W = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot (A_{1}-A_{2})\cdot \Delta s

The change is work loss in a year is:

\Delta W = (0.3)\cdot \left(\frac{1}{2}\right)\cdot (1.20\,\frac{kg}{m^{3}})\cdot (27.778\,\frac{m}{s})^{2}\cdot [(1.85\,m)\cdot (1.75\,m) - (1.50\,m)\cdot (1.75\,m)]\cdot (25\times 10^{6}\,m)

\Delta W = 2.043\times 10^{9}\,J

\Delta W = 2.043\times 10^{6}\,kJ

The change in chemical energy from gasoline is:

\Delta E = \frac{\Delta W}{\eta}

\Delta E = \frac{2.043\times 10^{6}\,kJ}{0.3}

\Delta E = 6.81\times 10^{6}\,kJ

The changes in gasoline consumption is:

\Delta m = \frac{\Delta E}{L_{c}}

\Delta m = \frac{6.81\times 10^{6}\,kJ}{44000\,\frac{kJ}{kg} }

\Delta m = 154.772\,kg

\Delta V = \frac{154.772\,kg}{0.74\,\frac{kg}{L} }

\Delta V = 209.151\,L

Lastly, the money saved is:

\Delta C = \left(\frac{154.772\,kg}{0.74\,\frac{kg}{L} }\right)\cdot (1.04\,\frac{USD}{L} )

\Delta C = 217.517\,USD

4 0
3 years ago
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