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3 years ago

Dan is applying for a job as an engineering technician. What are the tasks that he may be asked to perform?

Engineering

1 answer:

Vika [28.1K]3 years ago

5 0

Answer:

machine repair equipment maintenance

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Air at 105 kPa and 37° C flows upward through a6-cm-diameter inclined duct at a rate of 65 L/s. The duct diameter is then reduce

yaroslaw [1]

Answer:

0.129 m

Explanation:

See attached picture for step by step explanation.

5 0

3 years ago

25 points!!

adell [148]

Answer 1:

Like most electrical mechanical devices, electric motors have mechanical bearings that eventually wear out.

The graphite brushes and commutator of older DC motors can also wear out over time

Answer2:

Question two is not specific enough!!
Are we talking DC?? Because if so, then yes, nothing will happen. Assuming you are insulted by ground.
Ac is a different story, as the two would light up dangerously and etc.

-your welcome!!

5 0

4 years ago

Sea water with a density of 1025 kg/m3 flows steadily through a pump at 0.21 m3 /s. The pump inlet is 0.25 m in diameter. At the

myrzilka [38]

Answer:

$\dot W_{pump} = 16264.922\,W\,(16.265\,kW)$

Explanation:

The pump is modelled after applying Principle of Energy Conservation, whose form is:

$\frac{P_{1}}{\rho\cdot g}+ \frac{v_{1}^{2}}{2\cdot g} +z_{1} + h_{pump}=\frac{P_{2}}{\rho\cdot g}+ \frac{v_{2}^{2}}{2\cdot g} +z_{2}$

The head associated with the pump is cleared:

$h_{pump} = \frac{P_{2}-P_{1}}{\rho\cdot g}+\frac{v_{2}^{2}-v_{1}^{2}}{2\cdot g}+(z_{2}-z_{1})$

Inlet and outlet velocities are found:

$v_{1} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.25\,m)^{2} }$

$v_{1} \approx 4.278\,\frac{m}{s}$

$v_{2} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.152\,m)^{2} }$

$v_{2} \approx 11.573\,\frac{m}{s}$

Now, the head associated with the pump is finally computed:

$h_{pump} = \frac{175\,kPa-81.326\,kPa}{(1025\,\frac{kg}{m^{3}} )\cdot (9.807\,\frac{m}{s^{2}} )} +\frac{(11.573\,\frac{m}{s} )^{2}-(4.278\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )} + 1.8\,m$

$h_{pump} = 7.705\,m$

The power that pump adds to the fluid is:

$\dot W_{pump} = \dot V \cdot \rho \cdot g \cdot h_{pump}$

$\dot W_{pump} = (0.21\,m^{3})\cdot (1025\,\frac{kg}{m^{3}})\cdot (9.807\,\frac{m}{s^{2}})\cdot(7.705\,m)$

$\dot W_{pump} = 16264.922\,W\,(16.265\,kW)$

4 0

3 years ago

Why would an aerospace engineer limit the maximum angle of deflection of the control surfaces?

nordsb [41]

It has the same limit of each angle

4 0

3 years ago

To be safe, the engineers making the ride want to be sure the normal force does not exceed 1.8 times each persons weight - and t

Yuri [45]

Answer:

μ = 0.55

Explanation:

Given that

Normal weight = 1.8 x weight of person

N= 1.8 mg

We know that friction force Fr

Fr= μ N

μ=Coefficient of friction

N=Normal force

To find μ We have to equate friction and gravity force

Fr= Wt

μ N = m g

μ x 1.8 m g = m g

μ = 0.55

So the coefficient of friction will be 0.55.

5 0

3 years ago