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3 years ago

technician A says that voltage is electric pressure .Technician B says the amperage is electrical flow who is right

Engineering

1 answer:

Dmitrij [34]3 years ago

5 0

Answer:

I would say technician A

Explanation:

Definition of voltage- An electromotive force or potential difference expressed in volts.

electromotive force is pressure so technician A would make sense.

Read more about the perpendicular :

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6 0

2 years ago

A bridge to be fabricated of steel girders is designed to be 500 m long and 12 m wide at ambient temperature (assumed 20°C). Exp

Volgvan

Answer:

a) 22.5number

b) 22.22 m length

Explanation:

Given data:

Bridge length = 500 m

width of bridge = 12 m

Maximum temperature = 40 degree C

minimum temperature = - 35 degree C

Maximum expansion can be determined as

$\Delta L = L \alpha (T_{max} - T_{min})$

where , \alpha is expansion coefficient $= 12\times 10^{-6}$ degree C

SO, $\Delta L = 500\times 12\times 10^{-6}\times ( 40 - (-35))$

$\Delta L = 0.45 m = 450 mm$

number of minimum expansion joints is calculated as

$n = \frac{450}{20} = 22.5$

b) length of each bridge

$Length = \frac{500}{22.5} = 22.22 m$

8 0

3 years ago

A flow of 100 mgd is to be developed from a 190-mi^2 watershed. At the flow line the area's reservoir is estimated to cover 3900

yaroslaw [1]

Answer:

13-mi 27 acres

Explanation:

7 0

3 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

How might constraints and trade-offs affect the final design of a bridge?

Rainbow [258]

Answer:

It will limit the materials for your bridge which could cause damage to your bridge

Explanation:

4 0

3 years ago

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