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kvasek [131]
3 years ago
6

A wire is 0.92 m long and 1.2 mm2 in cross-sectional area. It carries a current of 5.0 A when a 2.2 V potential difference is ap

plied between its ends. Calculate the conductivity σ of the material of which this wire is made.

Engineering
1 answer:
Flura [38]3 years ago
3 0

The solution is in the attachment

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A saturated 1.5 ft3 clay sample has a natural water content of 25%, shrinkage limit (SL) of 12% and a specific gravity (GS) of 2
Svetllana [295]

79 f t^{3} is the volume of the sample when the water content is 10%.

<u>Explanation:</u>

Given Data:

V_{1}=100\ \mathrm{ft}^{3}

First has a natural water content of 25% = \frac{25}{100} = 0.25

Shrinkage limit, w_{1}=12 \%=\frac{12}{100}=0.12

G_{s}=2.70

We need to determine the volume of the sample when the water content is 10% (0.10). As we know,

V \propto[1+e]

\frac{V_{2}}{V_{1}}=\frac{1+e_{2}}{1+e_{1}}  ------> eq 1

e_{1}=\frac{w_{1} \times G_{s}}{S_{r}}

The above equation is at S_{r}=1,

e_{1}=w_{1} \times G_{s}

Applying the given values, we get

e_{1}=0.25 \times 2.70=0.675

Shrinkage limit is lowest water content

e_{2}=w_{2} \times G_{s}

Applying the given values, we get

e_{2}=0.12 \times 2.70=0.324

Applying the found values in eq 1, we get

\frac{V_{2}}{100}=\frac{1+0.324}{1+0.675}=\frac{1.324}{1.675}=0.7904

V_{2}=0.7904 \times 100=79\ \mathrm{ft}^{3}

7 0
3 years ago
If the rotational speed of a pump motor is reduced by 35%, what is the effect on the pump performance in terms of capacity, head
FinnZ [79.3K]

Answer:

- the capacity of the pump reduces by 35%.

- the head gets reduced by 57%.

the power consumption by the pump is reduced by 72%

Explanation:

the pump capacity is related to the speed as speed is reduces by 35%

so new speed is (100 - 35) = 65% of orginal speed

speed Q ∝ N ⇒ Q1/Q2 = N1/N2

Q2 = (N2/N1)Q1    

Q2 = (65/100)Q1

which means that the capacity of the pump is also reduces by 35%.

the head in a pump is related by

H ∝ N² ⇒ H1/H2 = N1²/N2²

H2 = (N2N1)²H1

H2 = (65/100)²H1 = 0.4225H1

so the head gets reduced by 1 - 0.4225 = 0.5775 which is 57%.

Now The power requirement of a pump is related as

P ∝ N³ ⇒ P1/P2 = N1³/N2³

P2 = (N2/N1)³P1

H2 = (65/100)²P1 = 0.274P1

So the reduction in power is 1 - 0.274 = 0.725 which is 72%

Therefore for a reduction of 35% of speed there is a reduction of 72% of the power consumption by the pump.  

8 0
2 years ago
Consider the expansion of a gas at a constant temperature in a water-cooled piston-cylinder system. The constant temperature is
Leona [35]

Answer:

Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})

Explanation:

According to the first thermodynamic law, the energy must be conserved so:

dQ = dU - dW

Where Q is the heat transmitted to the system, U is the internal energy and W is the work done by the system.

This equation can be solved by integration between an initial and a final state:

(1) \int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU - \int\limits^1_2 {} \, dW

As per work definition:

dW = F*dr

For pressure the force F equials the pressure multiplied by the area of the piston, and considering dx as the displacement:

dW = PA*dx

Here A*dx equals the differential volume of the piston, and considering that any increment in volume is a work done by the system, the sign is negative, so:

dW = - P*dV

So the third integral in equation (1) is:

\int\limits^1_2 {- P} \, dV

Considering the gas as ideal, the pressure can be calculated as P = \frac{n*R*T}{V}, so:

\int\limits^1_2 {- P} \, dV = \int\limits^1_2 {- \frac{n*R*T}{V}} \, dV

In this particular case as the systems is closed and the temperature constant, n, R and T are constants:

\int\limits^1_2 {- \frac{n*R*T}{V}} \, dV = -nRT \int\limits^1_2 {\frac{1}{V}} \, dV

Replacion this and solving equation (1) between state 1 and 2:

\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU + nRT \int\limits^1_2 {\frac{1}{V}} \, dV

Q_{2} - Q_{1} = U_{2} - U_{1} + nRT(ln V_{2} - ln V_{1})

Q_{2} - Q_{1} = U_{2} - U_{1} + nRT ln \frac{V_{2}}{V_{1}}

The internal energy depends only on the temperature of the gas, so there is no internal energy change U_{2} - U_{1} = 0, so the heat exchanged to the system equals the work done by the system:

Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})

4 0
3 years ago
Gas is kept in a 0.1 m diameter cylinder under the weight of a 100 kg piston that is held down by a spring with a stiffness k =
Artyom0805 [142]

Answer:

The spring is compressed by 0.275 meters.

Explanation:

For equilibrium of the gas and the piston the pressure exerted by the gas on the piston should be equal to the sum of  weight of the piston and the force the spring exerts on the piston

Mathematically we can write

Force_{pressure}=Force_{spring}+Weight_{piston}

we know that

Force_{pressure}=Pressure\times Area=300\times 10^{3}\times \frac{\pi \times 0.1^2}{4}=750\pi Newtons

Weight_{piston}=mass\times g=100\times 9.81=981Newtons

Now the force exerted by an spring compressed by a distance 'x' is given by Force_{spring}=k\cdot x=5\times 10^{3}\times x

Using the above quatities in the above relation we get

5\times 10^{3}\times x+981=750\pi \\\\\therefore x=\frac{750\pi -981}{5\times 10^{3}}=0.275meters

5 0
3 years ago
According to the video, what are examples of systems that Stationary Engineers oversee? Check all that apply. electrical systems
garik1379 [7]

Answer:

electrial systems

fire systems

heating systems

air systems

Explanation:

3 0
3 years ago
Read 2 more answers
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