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3 years ago

The two types of outlets that are found in an electrical system are:______

Engineering

1 answer:

Dmitry [639]3 years ago

3 0

Answer:

a. lighting and receptacle outlets

Explanation:

The two types of outlets that are found in an electrical system are

a. lighting and receptacle outlets

Outlets allow electrical equipment to connect to the electrical grid. The electrical grid provides alternating current to the outlet.

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A 20-mm-diameter steel bar is to be used as a torsion spring. If the torsional stress in the bar is not to exceed 110 MPa when o

ch4aika [34]

Answer:

1.887 m

Explanation:

(15 *pi)/180

= 0.2618 rad

Polar moment

= Pi*d⁴/32

= (22/7*20⁴)/32

= 15707.96

Torque on shaft

= ((22/7)*20³*110)/16

= 172857.14

= 172.8nm

Shear modulus

G = 79.3

L = Gjθ/T

= 79.3x10⁹x(1.571*10^-8)x0.2618/172.8

= 1.887 m

The length of the bar is therefore 1.887 meters

5 0

3 years ago

g A circular oil slick of uniform thickness is caused by a spill of one cubic meter of oil. The thickness of the oil slick is de

Anika [276]

Answer:

the rate of increase of radius is dR/dt = 0.804 m/hour = 80.4 cm/hour

Explanation:

the slick of oil can be modelled as a cylinder of radius R and thickness h, therefore the volume V is

V = πR² * h

thus

h = V / (πR²)

Considering that the volume of the slick remains constant, the rate of change of radius will be

dh/dt = V d[1/(πR²)]/dt

dh/dt = (V/π) (-2)/R³ *dR/dt

therefore

dR/dt = (-dh/dt)* (R³/2) * (π/V)

where dR/dt = rate of increase of the radius , (-dh/dt)= rate of decrease of thickness

when the radius is R=8 m , dR/dt is

dR/dt = (-dh/dt)* (R³/2) * (π/V) = 0.1 cm/hour *(8m)³/2 * π/1m³ *(1m/100 cm)= 0.804 m/hour = 80.4 cm/hour

4 0

3 years ago

The exhaust steam from a power station turbine is condensed in a condenser operating at 0.0738 bar(abs). The surface of the heat

lozanna [386]

Answer:

Percentage change 5.75 %.

Explanation:Given ;

Given

Pressure of condenser =0.0738 bar

Surface temperature=20°C

Now from steam table

Properties of steam at 0.0738 bar

Saturation temperature corresponding to saturation pressure =40°C

$h_f= 167.5\frac{KJ}{Kg},h_g= 2573.5\frac{KJ}{Kg}$

So Δh=2573.5-167.5=2406 KJ/kg

Enthalpy of condensation=2406 KJ/kg

So total heat=Sensible heat of liquid+Enthalpy of condensation

$Total\ heat\ =C_p\Delta T+\Delta h$

Total heat =4.2(40-20)+2406

Total heat=2,544 KJ/kg

Now film coefficient before inclusion of sensible heat

$h_1=\dfrac{\Delta h}{\Delta T}$

$h_1=\dfrac{2406}{20}$

$h_1=120.3\frac{KJ}{kg-m^2K}$

Now film coefficient after inclusion of sensible heat

$h_2=\dfrac{total\ heat}{\Delta T}$

$h_2=\dfrac{2,544}{20}$

$h_2=127.2\frac{KJ}{kg-m^2K}$

$So\ Percentage\ change=\dfrac{h_2-h_1}{h_1}\times 100$

$=\dfrac{127.2-120.3}{120.3}\times 100$

=5.75 %

So Percentage change 5.75 %.

3 0

3 years ago

The phasor technique is not valid if the frequencies of the sinusoids in the time domain are different. Part F - Use phasors to

AlladinOne [14]

Answer:

a, c

Explanation:

As the problem statement tells you, the phasor technique cannot be used when the frequencies are different. The frequencies are different when the coefficients of t are different. The different ones are highlighted.

a. 45 sin(2500t – 50°) + 20 cos(1500t +20°)

b. 25 cos(50t + 160°) + 15 cos(50t +70°)

c. 100 cos(500t +40°) + 50 sin(500t – 120°) – 120 cos(500t + 60°) -100 sin(10,000t +90°) + 40 sin(10, 100t – 80°) + 80 cos(10,000t)

d. 75 cos(8t+40°) + 75 sin(8t+10°) – 75 cos(8t + 160°)

4 0

3 years ago

Read 2 more answers

While servicing a vehicle, you retrieve the diagnostic code P0401, "EGR (exhaust gas recirculatior flow insufficient." Which of

n200080 [17]

Answer: Clogged EGR ports or passages

Explanation: The clogging causes insufficient flow.

3 0

2 years ago