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Harlamova29_29 [7]
3 years ago
9

5. One of the major road blocks for direct DC power in residences is the ___. A) generation of DC B) lack of standardization C)

difficulty of installation D) quantity of components required for this type of system
Engineering
1 answer:
Montano1993 [528]3 years ago
6 0

Answer:

D) quantity of components required for this type of system

Explanation:

Electricity can be transmitted using the alternating and direct currents. The alternating current is one in which the flow of the current diverts at certain time intervals whereas, the direct current is one in which there is a constant one-directional flow of current. The DC is used in batteries and solar panels. Residential areas and business places make use of the AC current.

One of the several reasons why the DC is not used in homes is because unlike the AC it is not easy to build and sustain. Moreso, it has more components compared to the AC. For example, its motor consists of brushes and commutators . The components, example, switches are also large compared to the AC components.

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cellular service/data

Explanation:

Mobile Data and hotspot

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Illustrate the crowbar protection for silicon controlled rectifier​
julsineya [31]

Answer:

The SCR over voltage crowbar or protection circuit is connected between the output of the power supply and ground. ... It also clamps the gate voltage at ground potential until the Zener turns on. The capacitor C1 is present to ensure that short spikes to not trigger the circuit.

4 0
3 years ago
What standard feature is on all circular saws?
PtichkaEL [24]

What standard feature is on all circular saws?

4. All of the above

4 0
3 years ago
A strip ofmetal is originally 1.2m long. Itis stretched in three steps: first to a length of 1.6m, then to 2.2 m, and finally to
andreyandreev [35.5K]

Answer:

strains for the respective cases are

0.287

0.318

0.127

and for the entire process 0.733

Explanation:

The formula for the true strain is given as:

\epsilon =\ln \frac{l}{l_{o}}

Where

\epsilon = True strain

l= length of the member after deformation

l_{o} = original length of the member

<u>Now for the first case we have</u>

l= 1.6m

l_{o} = 1.2m

thus,

\epsilon =\ln \frac{1.6}{1.2}

\epsilon =0.287

<u>similarly for the second case we have</u>

l= 2.2m

l_{o} = 1.6m   (as the length is changing from 1.6m in this case)

thus,

\epsilon =\ln \frac{2.2}{1.6}

\epsilon =0.318

<u>Now for the third case</u>

l= 2.5m

l_{o} = 2.2m

thus,

\epsilon =\ln \frac{2.5}{2.2}

\epsilon =0.127

<u>Now the true strain for the entire process</u>

l=2.5m

l_{o} = 1.2m

thus,

\epsilon =\ln \frac{2.5}{1.2}

\epsilon =0.733

6 0
3 years ago
A pump operating at steady state receives liquid water at 20°C, 100 kPa with a mass flow rate of 53 kg/min. The pressure of the
VARVARA [1.3K]

Answer:

Input Power = 6.341 KW

Explanation:

First, we need to calculate enthalpy of the water at inlet and exit state.

At inlet, water is at 20° C and 100 KPa. Under these conditions from saturated water table:

Since the water is in compresses liquid state and the data is not available in compressed liquid chart. Therefore, we use approximation:

h₁ = hf at 20° C = 83.915 KJ/kg

s₁ = sf at 20° C = 0.2965 KJ/kg.k

At the exit state,

P₂ = 5 M Pa

s₂ = s₁ = 0.2965 K J / kg.k    (Isentropic Process)

Since Sg at 5 M Pa is greater than s₂. Therefore, water is in compresses liquid state. Therefore, from compressed liquid property table:

h₂ = 88.94 KJ/kg

Now, the total work done by the pump can be calculated as:

Pump Work = W = (Mass Flow Rate)(h₂ - h₁)

W = (53 kg/min)(1 min/60 sec)(88.94 KJ/kg - 83.915 KJ/kg)

W = 4.438 KW

The efficiency of pump is given as:

efficiency = η = Pump Work/Input Power

Input Power = W/η

Input Power = 4.438 KW/0.7

<u>Input Power = 6.341 KW</u>

5 0
3 years ago
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