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kakasveta [241]

3 years ago

6

Mike is involved in developing the model building codes that various states and local authorities in the United States adopt. He

works with the __________ , which consists of members who are building code officials and building safety professionals.

1 answer:

Lyrx [107]3 years ago

8 0

<h3>Answer:</h3>

Mike is involved in developing the model building codes that various states and local authorities in the United States adopt. He works with the <u>Workers</u> , which consists of members who are building code officials and building safety professionals.

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A non-inductive load takes a current of 15A at 125V. An inductor is then connected in series in order that the same current shal

Norma-Jean [14]

Answer:

The inductance of the inductor is 0.051H

Explanation:

From Ohm's law;

V = IR .................. 1

The inductor has its internal resistance referred to as the inductive reactance, X $_{L}$ , which is the resistance to the flow of current through the inductor.

From equation 1;

V = IX $_{L}$

X $_{L}$ = $\frac{V}{I}$ ................ 2

Given that; V = 240V, f = 50Hz, $\pi$ = $\frac{22}{7}$ , I = 15A, so that;

From equation 2,

X $_{L}$ = $\frac{240}{15}$

= 16Ω

To determine the inductance of the inductor,

X $_{L}$ = 2 $\pi$ fL

L = $\frac{X_{L} }{2 \pi f}$

= $\frac{16}{2*\frac{22}{7}*50 }$

= 0.05091

The inductance of the inductor is 0.051H.

4 0

3 years ago

In a Rankine cycle, superheated steam that enters the turbine at 1273.15 K and 1.8 MPa is then expanded to a vapor at 0.1 MPa. W

GrogVix [38]

Answer:

The shaft work generated per kilogram is $-1.3 \frac{MJ}{kg}$

Explanation:

Given:

Temperature $T = 1273.15$ K

Initial Pressure $P_{1} = 1.8$ MPa

Final pressure $P_{2} = 0.1$ MPa

From the table superheated,

$h_{i} = 4635$ $\frac{K J}{Kg}$ and $h_{f} = 2706.54$ $\frac{K J}{Kg}$

Work done by shaft is,

$W = h_{f} - h_{i}$

$W = 2706.54 - 4635$

$W = -1928.46 \frac{kJ}{kg}$

But here efficiency is 0.56,

So work generated per kg is,

Work = $0.56 \times(- 1928.46)$

Work = $-1.3$ $\frac{MJ}{kg}$

Therefore, the shaft work generated per kilogram is $-1.3 \frac{MJ}{kg}$

6 0

3 years ago

Bài 3: Cho cơ cấu culít (hình 3.5) với các kích thước động lAB = 0,5lAC = 0,1m. Khâu 3 chịu tác dụng của mô men M3 = 500 N. Cơ c

lesya [120]

??????????????????????

3 0

3 years ago

A driver counts 21 other vehicles using 3 EB lanes on one section of I-80 between her rented car and an overpass ahead. It turne

SVEN [57.7K]

Answer:

vehicle density = 28.205 veh/mile

flow rate = 0.909 veh/hr

Explanation:

given data

count n = 21

distance = 0.78 miles

speed = 52 mph

solution

we get here vehicle density that is express as

vehicle density = n ÷ distance ...............1

vehicle density = ( 21 + 1 ) ÷ 0.78

vehicle density k = 28.205 veh/mile

and

now we get here flow rate that is express as

flow rate = k × vs .................2

flow rate = 28.205 × ( 52 × 0.00062 ÷ 1m )

flow rate = 0.909 veh/hr

6 0

3 years ago

Energy is generated uniformly in a 6 cm thick wall. The steady-state temperature distribution

mezya [45]

Answer:

Heat generation per unit volume is 45 KW.

Explanation:

Given that

Thickness of wall = 6 cm

Temperature distribution

$T(z)=145+3000z-1500z^2$ -----1

K= 15 W/m.k

As we know that at steady state condition

$\dfrac{d^2T}{dz^2}+\dfrac{q}{K}=0$ -----2

Where q is the heat generation per unit volume.

So from equation 1

$\dfrac{dT}{dz}=3000-3000z$

$\dfrac{d^2T}{dz^2}=-3000$

Now from equation 2

$\dfrac{d^2T}{dz^2}+\dfrac{q}{K}=0$

$-3000+\dfrac{q}{15}=0$

So q= 45 KW

So heat generation per unit volume is 45 KW.

6 0

3 years ago