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BlackZzzverrR [31]

4 years ago

12

For installation with a 25-kVA, 3-phase transformer, a 440-volt primary, and a 120-volt secondary. Calculate the maximum overcur

rent protection value (not the fuse or circuit breaker rating) using the protection method for primary-only protection. (Round the FINAL answer to two decimal places.)

1 answer:

andre [41]4 years ago

4 0

Answer:

41.053 A

Explanation:

given,

three phase kVA = 25-kVA

voltage = 440 Volt

current = ?

To determine three phase kVA when volts and amperes are know

three phase kVA = 1.73 x V x I

$I = \dfrac{three\ phase\ kVA}{1.73 \times V}$

$I = \dfrac{25000}{1.73 \times 440}$

I = 32.84 A

the maximum overcurrent protection value is equal to

= 125 % of I

= 1.25 x 32.84

= 41.053 A

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Answer:

C. The initial momentum should be equal to the final momentum due to the conservation of momentum.

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3 years ago

A thin oil slick (no=1.50) floats on water (nw=1.33). When a beam of white light strikes this film at normal incidence from air,

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Answer:

The (minimum) thickness of the oil slick is 325 nm.

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similarly,

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Therefore:

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3 years ago

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3 years ago

Read 2 more answers

The Earth is 1.5 × 1011 m from the Sun and takes a year to make one complete orbit. It rotates on its own axis once per day. It

katrin2010 [14]

Answer:

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$v_{cm} = 2.98 \times 10^4 m/s$

Part b)

$K_{trans} = 2.68 \times 10^{33} J$

Part c)

$\omega = 7.27 \times 10^{-5} rad/s$

Part d)

$KE_{rot} = 2.6 \times 10^{29} J$

Part e)

$KE_{tot} = 2.68 \times 10^{33} J$

Explanation:

Time period of Earth about Sun is 1 Year

so it is

$T = 1 year = 3.15 \times 10^7 s$

now we know that angular speed of the Earth about Sun is given as

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{3.15 \times 10^7}$

now speed of center of Earth is given as

$v_{cm} = r\omega$

$r = 1.5 \times 10^{11} m$

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$v_{cm} = 2.98 \times 10^4 m/s$

Part b)

now transnational kinetic energy of center of Earth is given as

$K_{trans} = \frac{1}{2}mv^2$

$K_{trans} = \frac{1}{2}(6 \times 10^{24})(2.98 \times 10^4)^2$

$K_{trans} = 2.68 \times 10^{33} J$

Part c)

Angular speed of Earth about its own axis is given as

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{24 \times 3600}$

$\omega = 7.27 \times 10^{-5} rad/s$

Part d)

Now moment of inertia of Earth about its own axis

$I = \frac{2}{5}mR^2$

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$I = 9.83 \times 10^{37} kg m^2$

now rotational energy is given as

$KE_{rot} = \frac{1}{2}I\omega^2$

$KE_{rot} = \frac{1}{2}(9.83 \times 10^{37})(7.27 \times 10^{-5})^2$

$KE_{rot} = 2.6 \times 10^{29} J$

Part e)

Now total kinetic energy is given as

$KE_{tot} = KE_{trans} + KE_{rot}$

$KE_{tot} = 2.68 \times 10^{33} + 2.6 \times 10^{29}$

$KE_{tot} = 2.68 \times 10^{33} J$

6 0

3 years ago