What engineers call moment, scientists call

Technician A says that if fuel pump pressure is correct, fuel pump volume will be correct as well. Technician B says that a fuel

guajiro [1.7K]

Answer:

Technician B only

Explanation:

hope this helps :)

5 0

2 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

The 1000-lb elevator is hoisted by the pulley system and motor M. The motor exerts a constant force of 500 lb on the cable. The

klemol [59]

The power that must be supplied to the motor is 136 hp

<u>Explanation:</u>

Given-

weight of the elevator, m = 1000 lb

Force on the table, F = 500 lb

Distance, s = 27 ft

Efficiency, ε = 0.65

Power = ?

According to the equation of motion:

F = ma

$3(500) - 1000 = \frac{1000}{32.2} * a$

a = 16.1 ft/s²

We know,

$v^2 - u^2 = 2a (S - So)\\\\v^2 - (0)^2 = 2 * 16.1 (27-0)\\\\v = 29.48m/s$

To calculate the output power:

Pout = F. v

Pout = 3 (500) * 29.48

Pout = 44220 lb.ft/s

As efficiency is given and output power is known, we can calculate the input power.

ε = Pout / Pin

0.65 = 44220 / Pin

Pin = 68030.8 lb.ft/s

Pin = 68030.8 / 500 hp

= 136 hp

Therefore, the power that must be supplied to the motor is 136 hp

5 0

3 years ago

engineering uses data from pareto charts to analyse motor faults caused during their production. Explain one advantage of using

lions [1.4K]

The advantage of a pareto chart is to make sure they have all of their tools

3 0

3 years ago

If you are a mechanical engineer answer these questions:

Natasha_Volkova [10]

Answer:

1. Yes, they are all necessary.

2. Both written and verbal communication skills are of the utmost importance in business, especially in engineering. Communication skills boost you or your teams' performance because they provide clear information and expectations to help manage and deliver excellent work.

3 0

2 years ago