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1 year ago

12

An installation consists of a 30-kVA, 3-phase transformer, a 480-volt primary, and a 240-volt secondary. Calculate the largest s

tandard size circuit breaker permitted for primary-only protection to be used without applying Note 1 of Table 450.3(B).

1 answer:

Nikitich [7]1 year ago

5 0

Answer: 45 A

Explanation:

Primary only protection 3-phase

I = 3 phase kVA / ( 1.723 * V)

I = 30000 / ( 1.732 * 480 ) = 36.085 A

Table 450.3(B)

Currents of 9A or more column

primary only protection = 125%

Max OCPD pri = 125% of I = 1.25 * 36.085 = 45.11 A

Table 450.3(B) Note 1 does not apply, use next smaller Table 240.6(A)

Next smaller = 45 A

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What s physics?

Alex777 [14]

The answer is (B. The study of Matter and Energy) but technically you could consider physics all of these as engineering is based on physics and that would be the study of inventions, chemistry and biology were both discovered because of physics, and physics invokes more math than any other subject as it applies math to the entire Universe.

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4 years ago

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In a Young's double-slit experiment, 610-nm-wavelength light is sent through the slits. The intensity at an angle of 2.95° from

Sliva [168]

Answer:

spacing between the slits is 405.32043 × $10^{-9}$ m

Explanation:

Given data

wavelength = 610 nm

angle = 2.95°

central bright fringe = 85%

to find out

spacing between the slits

solution

we know that spacing between slit is

I = 4 $I_{0}$ × cos²∅/2

so

I/4 $I_{0}$ = cos²∅/2

here I/4 $I_{0}$ is 85 % = 0.85

so

0.85 = cos²∅/2

cos∅/2 = √0.85

∅ = 2 × $cos^{-1}$ 0.921954

∅ = 45.56°

∅ = 45.56° ×π/180 = 0.7949 rad

and we know that here

∅ = 2π d sinθ / wavelength

so

d = ∅× wavelength / ( 2π sinθ )

put all value

d = 0.795 × 610× $10^{-9}$ / ( 2π sin2.95 )

d = 405.32043 × $10^{-9}$ m

spacing between the slits is 405.32043 × $10^{-9}$ m

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3 years ago

a horse began running due east and covered 25km in 4.0 hours what is the average velocity of the horse

Komok [63]

25km / 4hr = <em>6.25 km/hr east</em>

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3 years ago

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A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a lar

kakasveta [241]

Answer:

$W=4037.36\ J$

Explanation:

Given:

mass of ice melted, $m=3.3\times10^{-2}\ kg$

time taken by the ice to melt, $t=5\ min=300\ s$

latent heat of the ice, $L=3.34\times 10^5\ J$

Now the heat rejected by the Carnot engine:

$Q_R=m.L$

$Q_R=0.033\times 3.34\times 10^5$

$Q_R=11022\ J$

Since we have boiling water as hot reservoir so:

$T_H=373\ K$

The cold reservoir is ice, so:

$T_L=273\ K$

Now the efficiency:

$\eta=1-\frac{T_L}{T_H}$

$\eta=1-\frac{273}{373}$

$\eta=26.81\%$

Now form the law of energy conservation:

Heat supplied:

$Q_S-W=Q_R$

where:

$Q_S=$ heat supplied to the engine

$Q_S-\eta\times Q_S=Q_R$

$Q_S(1-\eta)=Q_R$

$Q_S=\frac{11022}{1-0.2681}$

$Q_S=15059.36\ J$

Now the work done:

$W=Q_S-Q_R$

$W=15059.36-11022$

$W=4037.36\ J$

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3 years ago

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C .Clastic,why you see (clastic called Detrial ,made of broken rocks so the answer is C

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3 years ago