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2 years ago

Tony works as a Sorter in a processing factory. Which qualifications does he most likely have?

Engineering

2 answers:

svp [43]2 years ago

7 0

Answer:

answer is B

Explanation:

Vlada [557]2 years ago

4 0

Answer:

its b

Explanation:

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Suppose you have two boxes in front of you. One box contains a Thevenin Equivalent (voltage source in series with a resistor) an

fomenos

Answer:

1. Measure the temperature of the boxes and leave them unconnected.

2. Norton reduces his circuit down to a single resistance in parallel with a constant current source. A real-life Norton equivalent circuit would be continuously wasting power (as heat) as the current source dumps energy into the resistor, even when externally unconnected, while a Thevenin equivalent circuit would sit there doing nothing.

3. The Norton equivalent box would get warm and eventually run out of power. The Thevenin equivalent box would stay at ambient temperature.

8 0

3 years ago

A gas enters a compressor that provides a pressure ratio (exit pressure to inlet pressure) equal to 8. If a gage indicates the g

olga55 [171]

Answer:

$P_2_{abs}=160\ psia$ (absolute).

Explanation:

Given that

Pressure ratio r

r=8

$r=\dfrac{P_2_{abs}}{P_1_{abs}}$

$8=\dfrac{P_2_{abs}}{P_1_{abs}}$ -----1

P₁(gauge) = 5.5 psig

We know that

Absolute pressure = Atmospheric pressure + Gauge pressure

Given that

Atmospheric pressure = 14.5 lbf/in²

P₁(abs) = 14.5 + 5.5 psia

P₁(abs) =20 psia

Now by putting the values in the above equation 1

$8=\dfrac{P_2_{abs}}{20}$

$P_2_{abs}=8\times 20\ psia$

$P_2_{abs}=160\ psia$

Therefore the exit gas pressure will be 160 psia (absolute).

7 0

3 years ago

Technician A says that weld-on dent removal attachments may be used on a steel panel.

spin [16.1K]

Answer:

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Explanation:

7 0

2 years ago

Which factor has the most significant impact in design?

wariber [46]

D) cost

The fund available for a project is the first factor to be considered to ensure that the project comes to reality and serve it’s purpose

3 0

1 year ago

A fluid with a relative density of 0.9 flows in a pipe which is 12 m long and lies at an angle of 60° to the horizontal At the t

Minchanka [31]

Answer:

$Q=7.3\times 10^{-3} m^3/s$

Explanation:

Given that

At top $d_2=30 mm,P_2=860 KPa ,P_1=1000 KPa,d_1=85 mm$

$\rho =900\dfrac{Kg}{m^3}$

We know that

$\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2$

$A_1V_1=A_2V_2$

$\frac{V_1}{V_2}=\left(\dfrac{d_2}{d_1}\right)^2$

$\frac{V_1}{V_2}=\left(\dfrac{30}{85}\right)^2$

$V_2=8.02V_1$

$Z_2=12 sin60^{\circ}$

$\dfrac{1000\times 1000}{900\times 9.81}+\dfrac{V_1^2}{2\times 9.81}+0=\dfrac{860\times 1000}{900\times 9.81 }+\dfrac{V_2^2}{2\times 9.81}+12 sin60^{\circ}$

So $V_1=1.30$ m/s

We know that flow rate Q=AV

$Q=A_1V_1$

By putting the values

$A_1=\dfrac{\pi}{4}d^2$

$Q=7.3\times 10^{-3} m^3/s$

To find the flow rate we do not need the direction of flow,because we are just doing balancing of energy at inlet and at the exits of pipe.

4 0

3 years ago