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Svetlanka [38]
3 years ago
10

Electrical metallic tubing would be used in an electrical installation because it

Physics
2 answers:
maw [93]3 years ago
6 0
Hey There!

Electrical metallic tubing would be used in an electrical installation because it <span>is less expensive.</span>
Pani-rosa [81]3 years ago
5 0
<h3><u>Answer;</u></h3>

is less expensive than rigid metal conduit.

Electrical metallic tubing would be used in an electrical installation because<em><u> it is less expensive than rigid metal conduit.</u></em>

<h3><u>Explanation</u>;</h3>
  • Electrical conduit refers to a tube that is used to protect and route electrical wiring in a building or structure.
  • Electrical conduits are made of plastics, metals or even fiber.
  • <em><u>Electrical metallic tubing or EMT, is commonly used instead of galvanized rigid conduit since it is less expensive and also lighter than the galvanized rigid conduit.</u></em>
  • <em><u>Electrical metallic conduit is more commonly used in commercial buildings and industrial buildings as compared to residential buildings.</u></em>
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A jet airliner moving initially at 406 mph (with respect to the ground) to the east moves into a region where the wind is blowin
astraxan [27]

Answer:

966 mph

Explanation:

Using as convention:

- East --> positive x-direction

- North --> Positive y-direction

The x- and y- components of the initial velocity of the jet can be written as

v_{1x} = 406 mph\\v_{1y} = 0

While the components of the velocity of the wind are

v_{2x} = (568)(cos 15^{\circ})=548.6 mph\\v_{2y} = (568)(sin 15^{\circ})=147.0 mph

So the components of the resultant velocity of the jet are

v_x = v_{1x}+v_{2x}=406+548.6=954.6 mph\\v_y = v_{1y}+v_{2y}=0+147.0=147.0 mph

And the new speed is the magnitude of the resultant velocity:

v=\sqrt{v_x^2+v_y^2}=\sqrt{(954.6)^2+(147.0)^2}=965.8 mph \sim 966 mph

6 0
3 years ago
Which of these nebulae is the odd one out?
Lubov Fominskaja [6]

Answer: The answer is D!

Explanation:

5 0
3 years ago
Read 2 more answers
A periodic wave with wavelength 2m has a speed of 4m/s. What is the waves frequency?.
Gekata [30.6K]

The wave frequency is 2 Hz.

What is wave frequency ?

The number of waves that pass through a fixed point in a given amount of time is referred to as the wave frequency. The hertz is the SI unit for wave frequency (Hz).

f = v / w

where,

f = frequency\\v = speed \\w = wavelength

Given,

v = 4 m/s, w = 2 m

f = 4/2 \\f = 2 Hz\\

The waves frequency is 2 Hz.

To know more about wave frequency,check out:

brainly.com/question/15830195

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5 0
1 year ago
A metal wire is in thermal contact with two heat reservoirs at both of its ends. Reservoir 1 is at a temperature of 563 K, and r
Schach [20]

Answer:

1.85 J/K

Explanation:

The computation of total change in entropy is shown below:-

Change in Entropy = Sum Q ÷ T

= \frac{-heat\ entering\ the\ reservoir}{Reservoir\ 1\ Temperature} + \frac{Conduction\ of\ heat}{Reservoir\ 2\ Temperature}

= \frac{-1760}{563} + \frac{1760}{354}

= -3.12 + 4.97

= 1.85 J/K

Therefore for computing the total change in entropy we simply applied the above formula.

As we can see that there is heat entering the reservoir so it will be negative while cold reservoir will be positive else the process would be impossible.

8 0
3 years ago
61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A
Mademuasel [1]

Answer:

Part a)

percentage = 21.3%

Part b)

percentage = 2.13 \times 10^{-5}%

Explanation:

As we know that total power used in the room is given as

P = P_1 + P_2 + P_3 + P_4

here we have

P_1 = (110)(3) = 330 W

P_2 = 100 W

P_3 = 60 W

P_4 = 3 W

P = 330 + 100 + 60 + 3

P = 493 W

Part a)

Since power supply is at 110 Volt so the current obtained from this supply is given as

110\times i = 493

i = 4.48 A

now resistance of transmission line

R = \frac{\rho L}{A}

R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}

R = 5.23 \ohm

now power loss in line is given as

P = i^2 R

P = (4.48)^2(5.23)

P = 105 W

Now percentage loss is given as

percentage = \frac{loss}{supply} \times 100

percentage = \frac{105}{493} \times 100

percentage = 21.3%

Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

110 \times 10^3 (i ) = 493

i = 4.48\times 10^{-3} A

now power loss in line is given as

P = i^2 R

P = (4.48 \times 10^{-3})^2(5.23)

P = 1.05 \times 10^{-4} W

Now percentage loss is given as

percentage = \frac{loss}{supply} \times 100

percentage = \frac{1.05 \times 10^{-4}}{493} \times 100

percentage = 2.13 \times 10^{-5}%

6 0
3 years ago
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