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3 years ago

In a typical American building, most modern lighting systems must use what voltage?

Engineering

1 answer:

kakasveta [241]3 years ago

7 0

Answer:

Typically, either 110-volt AC (110V) or 220-volt AC (220V) is used. Most countries use 50Hz (50 Hertz or 50 cycles per second) as their AC frequency. Only a handful use 60Hz. The standard in the United States is 120V and 60Hz AC electricity.

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Writing an excellent problem statement will not help guide you through the rest of the process and steer you towards the BEST so

Ilya [14]

Answer:

Writing an excellent problem statement will not help guide you through the rest of the process and steer you towards the BEST solution.

False

Explanation:

An excellent problem statement sets the overall tone for the rest of the engineering process, whether it be at the analysis, design, or implementation stages. This is why a problem statement must be focused, clear, and specific. An excellent problem statement contains the problem definition, method for solving the problem (the claim proposed), purpose, statement of objectives, and scope. For an excellent problem statement to be effective, it must also show the gap that is to be closed to achieve the intended objective.

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3 years ago

A wastewater plant discharges a treated effluent (w) with a flow rate of 1.1 m^3/s, 50 mg/L BOD5 and 2 mg/L DO into a river (s)

4vir4ik [10]

A wastewater plant discharges a treated effluent (w) with a flow rate of 1.1 m^3/s, 50 mg/L BOD5 and 2 mg/L DO into a river (s) with a flow rate of 8.7 m^3/s, 6 mg/L BOD5 and 8.3 mg/L DO. Both streams are at 20Â°C. After mixing, the river is 3 meters deep and flowing at a velocity of 0.50 m/s. DOsat for this river is 9.0 mg/L. The deoxygenation constant is kd= 0.20 d^-1 and The reaction rate constant k at 20 Â°C is 0.27 d^-1.

The answer therefore would be the number 0.27 divided by two and then square while getting the square you would make it a binomial.

I wont give the answer but the steps

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3 years ago

Route Choice The cost of roadway improvements to the developer is a function of the amount of traffic being generated by the the

aksik [14]

Row choice the cost of roadway improvements to the developer and functional the amount of trafficBeing generated by the theater as well as the Ralph’s ladies

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3 years ago

5 cause of a electrical problem

cupoosta [38]

Answer:

1. Poor circuit protection

2.Grounding issue

3. lighting problem

4. Electrical shocks

5. High electricity bills

Explanation:

8 0

3 years ago

P10.12. A certain amplifier has an open-circuit voltage gain of unity, an input resistance of and an output resistance of The si

klio [65]

complete question

A certain amplifier has an open-circuit voltage gain of unity, an input resistance of 1 \mathrm{M} \Omega1MΩ and an output resistance of 100 \Omega100Ω The signal source has an internal voltage of 5 V rms and an internal resistance of 100 \mathrm{k} \Omega.100kΩ. The load resistance is 50 \Omega.50Ω. If the signal source is connected to the amplifier input terminals and the load is connected to the output terminals, find the voltage across the load and the power delivered to the load. Next, consider connecting the load directly across the signal source without the amplifier, and again find the load voltage and power. Compare the results. What do you conclude about the usefulness of a unity-gain amplifier in delivering signal power to a load?

Answer:

3.03 V 0.184 W

2.499 mV 125*10^-9 W

Explanation:

First, apply voltage-divider principle to the input circuit: 1

$V_{i}= (R_i/R_i+R_s) *V_s = 10^6/10^6+(0.1*10^6)\\$ *5

= 4.545 V

The voltage produced by the voltage-controlled source is:

A_voc*V_i = 4.545 V

We can find voltage across the load, again by using voltage-divider principle:

V_o = A_voc*V_i*(R_o/R_l+R_o)

= 4.545*(100/100+50)

= 3.03 V

Now we can determine delivered power:

P_L = V_o^2/R_L

= 0.184 W

Apply voltage-divider principle to the circuit:

V_o = (R_o/R_o+R_s)*V_s

= 50/50+100*10^3*5

= 2.499 mV

Now we can determine delivered power:

P_l = V_o^2/R_l

= 125*10^-9 W

Delivered power to the load is significantly higher in case when we used amplifier, so a unity gain amplifier can be useful in situation when we want to deliver more power to the load. It is the same case with the voltage, no matter that we used amplifier with voltage open-circuit gain of unity.

4 0

3 years ago