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Whitepunk [10]
3 years ago
7

True or False: Galvanized steel pipe can withstand moderate freezing.

Engineering
1 answer:
yaroslaw [1]3 years ago
5 0

Answer:

True

Explanation:

Older galvanized steel pipes, which have a tendency to freeze, are a bit more forgiving and will likely not burst. They can withstand extreme cold and warm temperatures.

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Consider the circuit below where R1 = R4 = 5 Ohms, R2 = R3 = 10 Ohms, Vs1 = 9V, and Vs2 = 6V. Use superposition to solve for the
VladimirAG [237]

Answer:

The value of v2 in each case is:

A) V2=3v for only Vs1

B) V2=2v for only Vs2

C) V2=5v for both Vs1 and Vs2

Explanation:

In the attached graphic we draw the currents in the circuit. If we consider only one of the batteries, we can consider the other shorted.

Also, what the problem asks is the value V2 in each case, where:

V_2=I_2R_2=V_{ab}

If we use superposition, we passivate a battery and consider the circuit affected only by the other battery.

In the first case we can use an equivalent resistance between R2 and R3:

V_{ab}'=I_1'R_{2||3}=I_1'\cdot(\frac{1}{R_2}+\frac{1}{R_3})^{-1}

And

V_{S1}-I_1'R_1-I_1'R_4-I_1'R_{2||3}=0 \rightarrow I_1'=0.6A

V_{ab}'=I_1'R_{2||3}=3V=V_{2}'

In the second case we can use an equivalent resistance between R2 and (R1+R4):

V_{ab}''=I_3'R_{2||1-4}=I_3'\cdot(\frac{1}{R_2}+\frac{1}{R_1+R_4})^{-1}

And

V_{S2}-I_3'R_3-I_3'R_{2||1-4}=0 \rightarrow I_3'=0.4A

V_{ab}''=I_3'R_{2||1-4}=2V

If we consider both batteries:

V_2=I_2R_2=V_{ab}=V_{ab}'+V_{ab}''=5V

7 0
3 years ago
g (d) Usually, in the case of two finite-duration signals like in parts (a) and (b), the convolution integralmust be evaluated i
Bas_tet [7]

you face is A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.

3 0
3 years ago
Consider a plane composite wall that is composed of two materials of thermal conductivities kA = 0.1 W/m*K and kB = 0.04 W/m*K a
nadya68 [22]

Answer:

q=39.15 W/m²

Explanation:

We know that

Thermal resistance due to conductivity given as

R=L/KA

Thermal resistance due to heat transfer coefficient given as

R=1/hA

Total thermal resistance

R_{th}=\dfrac{L_A}{AK_A}+\dfrac{L_B}{AK_B}+\dfrac{1}{Ah_1}+\dfrac{1}{Ah_2}+\dfrac{1}{Ah_3}

Now by putting the values

R_{th}=\dfrac{0.01}{0.1A}+\dfrac{0.02}{0.04A}+\dfrac{1}{10A}+\dfrac{1}{20A}+\dfrac{1}{0.3A}

R_{th}=4.083/A\ K/W

We know that

Q=ΔT/R

Q=\dfrac{\Delta T}{R_{th}}

Q=A\times \dfrac{200-40}{4.086}

So heat transfer per unit volume is 39.15 W/m²

q=39.15 W/m²

4 0
3 years ago
Which of the following is an activity of daily living? jogging cleaning weightlifting all of the above
Mrrafil [7]

Answer:

cleaning

Explanation:

4 0
3 years ago
John wants to construct a device using quartz crystal, Which device can he construct?
tatiyna

Answer: Option D, piezoelectric pressure guage

Explanation: Quartz crystal possess a very useful quality in science as they can generate small charges when pressure is applied to them or when they are hit. This property can be harnessed to construct a piezoelectric pressure gauge which would be used to measure and indicate changes in pressure, the quartz crystal releases little voltage each time there is an applied pressure . This device would be able to sense changes in pressure as there would voltage proportional to the applied pressure.

4 0
3 years ago
Read 2 more answers
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