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4 years ago

Truth table for (A+B)(B+C)

Engineering

1 answer:

serg [7]4 years ago

5 0

Answer:

A truth table is a logical expression of the given inputs specially in boolean algebra and boolean functions in computer science and digital electronics.

Explanation:

Number of variables are 3 i.e A,B and C

so the number of bits are 2^3=8

A B C A+B B+C (A+B)(B+C)

0 0 0 0 0 0

0 0 1 0 1 0

0 1 0 1 1 1

0 1 1 1 1 1

1 0 0 1 0 0

1 0 1 1 1 1

1 1 0 1 1 1

1 1 1 1 1 1

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Technician A says that a 12 Volt light bulb that draws 12 amps has a power output of 1 watt. Technician B says that a motor that

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Answer:

Technician B

Explanation:

Resistance, $R=\frac {V}{I}$ where V is the voltage and I is the current in amps

Therefore, $R=\frac {12}{12}=1 ohm$

Power=VI=12*12=144 W

Therefore, the power is 144 W and resistance is 1 Ohm. This implies that technician A is wrong while technician B is correct

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4 years ago

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Answer:

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3 years ago

Compared to arc welding, which of the following statements are true about gas welding?

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3 years ago

Air is contained in a cylinder device fitted with a piston-cylinder. The piston initially rests on a set of stops, and a pressur

Veseljchak [2.6K]

Answer:

The amount of heat transferred to the air is 340.24 kJ

Explanation:

From P-V diagram,

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final pressure P3 = P2 = 300 kPa

volume at point 2, V2 = V1 = 0.4 m³

final temperature T2 = T3 = 1200 K

To determine the final pressure V3, use ideal gas equation

PV = mRT

Where R is the specific gas constant = 0.2870 KPa m³ kg K

But,

from initial condition, mass m = PV/RT

m = (P1*V1)/R*T1

T1 = 27+273 = 300K

m = (100*0.4)/(0.2870*300) = 0.4646 kg

Then;

Final volume V3 = mRT3/P3

V3 = (0.4646*0.2870*1200)/300

V3 = 0.5334 m³

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W = P3*(V3-V2)

W = 300*(0.5334-0.4) = 40.02 kJ

To get the internal energy, the heat capacity at room temperature Cv is 0.718 kJ/kg K

∆U = m*Cv*(T2-T1)

∆U = 0.4646*0.718(1200-300)

∆U = 300.22 kJ

The heat transfer Q = W + ∆U

Q = 40.02 + 300.22 = 340.24 kJ

Determine the amount of heat transferred to the air, in kJ, while increasing the temperature to 1200 K is 340.24 kJ

The attached file shows the Pressure - Volume relationship (P -V graph)

7 0

4 years ago

Consider a voltage v = Vdc + vac where Vdc = a constant and the average value of vac = 0. Apply the integral definition of RMS t

Anna11 [10]

Answer:

Proof is as follows

Proof:

Given that , $V = V_{ac} + V_{dc}$

for any function f with period T, RMS is given by

$RMS = \sqrt{\frac{1}{T}\int\limits^T_0 {[f(t)]^{2} } \, dt }$

In our case, function is $V = V_{ac} + V_{dc}$

$RMS = \sqrt{\frac{1}{T}\int\limits^T_0 {[V_{ac} + V_{dc}]^{2} } \, dt }$

Now open the square term as follows

$RMS = \sqrt{\frac{1}{T}\int\limits^T_0 {[V_{ac}^{2} + V_{dc}^{2} + 2V_{dc}V_{ac}] } \, dt }$

Rearranging terms

$RMS = \sqrt{\frac{1}{T}\int\limits^T_0 {V_{dc}^{2} } \, dt + \frac{1}{T}\int\limits^T_0 {V_{ac}^{2} } \, dt + \frac{1}{T}\int\limits^T_0 {2V_{dc}V_{ac} } \, dt }$

You can see that

second term is square of RMS value of Vac

Third terms is average of VdcVac and given is that average of $V_{ac}V_{dc} = 0$

$RMS = \sqrt{\frac{1}{T}TV_{dc}^{2} + [RMS~~ of~~ V_{ac}]^2 }$

$RMS = \sqrt{V_{dc}^{2} + [RMS~~ of~~ V_{ac}]^2 }$

So it has been proved that given expression for root mean square (RMS) is valid

7 0

3 years ago