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lawyer [7]
3 years ago
8

An installation consists of a 10-kVA, single-phase transformer with a 440-volt primary and a 110-volt, 2-wire secondary using in

sulation type THWN copper conductors. No secondary OCPD is installed. The secondary loads are considered to be noncontinuous loads. Calculate the minimum size secondary conductors required to handle the secondary full load current for this transformer.
Physics
1 answer:
Setler79 [48]3 years ago
6 0

Answer:

<em>Use a 150-amp breaker and No. 1/0 copper conductors</em>

Explanation:

An installation consists of a 10-kVA, single-phase transformer with a 440-volt primary and a 110-volt, 2-wire secondary using insulation type THWN copper conductors. No secondary OCPD is installed. The secondary loads are considered to be noncontinuous loads. Calculate the minimum size secondary conductors required to handle the secondary full load current for this transformer

Calculate primary full-load amps:

10 kVA ÷ (440 volts ÷ 1000) = 22.72 amps

Calculate required feeder breaker size and conductor ampacity:

22.72 amps x 1.5 = 34.09 amps

Use an 80-amp breaker and No. 3 AWG copper conductors 1

Calculate secondary full-load amps:

10 kVA ÷ (110 volts ÷ 1000) = 90.9 amps

Calculate required secondary breaker size and conductor ampacity:

90.9 amps x 1.25 = 113.6 amps

Use a 150-amp breaker and No. 1/0 THWN copper conductors.

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3 years ago
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A stunt driver rounds a banked, circular curve. The driver rounds the curve at a high, constant speed, such that the car is just
bagirrra123 [75]

Answer: C

Frictional force

Explanation:

The description of the question above is an example of a circular motion.

For a car travelling in a curved path, the frictional force between the tyres and the road surface will provide the centripetal force.

Since the road is banked, and the cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car, for cornering the banked road, the car will not rely only on the frictional force.

Therefore, the correct answer is option C - the frictional force.

5 0
3 years ago
A swimming pool, 10.0 m by 4.0 m, is filled with water to a depth of 3.0 m at a temperature of 20.2°c. if the energy needed to r
Tomtit [17]

You need to find the mass of water in the pool.
Find the volume (10 x 4 x 3) = 120 m3

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3 years ago
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Hanna tosses a ball straight up with enough speed to remain in the air for several seconds?
olga_2 [115]
A) the velocity is 0 m/s
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3 years ago
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Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du
Klio2033 [76]

a. The disk starts at rest, so its angular displacement at time t is

\theta=\dfrac\alpha2t^2

It rotates 44.5 rad in this time, so we have

44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}

b. Since acceleration is constant, the average angular velocity is

\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2

where \omega_f is the angular velocity achieved after 6.00 s. The velocity of the disk at time t is

\omega=\alpha t

so we have

\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}

making the average velocity

\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}

Another way to find the average velocity is to compute it directly via

\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}

c. We already found this using the first method in part (b),

\omega=14.8\dfrac{\rm rad}{\rm s}

d. We already know

\theta=\dfrac\alpha2t^2

so this is just a matter of plugging in t=12.0\,\mathrm s. We get

\theta=179\,\mathrm{rad}

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2

Then for t=6.00\,\rm s we would get the same \theta=179\,\rm rad.

7 0
3 years ago
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