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2 years ago

What the minimum wire size for a general residential application on a 20 A circuit

Engineering

1 answer:

prisoha [69]2 years ago

7 0

Answer:

The minimum wire size for a 20 A circuit would be 14 gauge wire. However modern fire and building (in the United States) code dictates that 12 gauge wire be used for 20 A. This is for added safety because there is a chance that 14 gauge wire can be overloaded by 20 A if the situation is right.

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Why do many sources of water need treatment

Zarrin [17]

Answer:Even though EPA regulates and sets standards for public drinking water, many Americans use a home water treatment unit to: Remove specific contaminants. Take extra precautions because a household member has a compromised immune system. Improve the taste of drinking water.

3 0

3 years ago

#198. Moment of inertia about center of a segmented bar A bar of width is formed of three uniform segments with lengths and area

zaharov [31]

Complete Complete

The complete question is shown on the first uploaded image

Answer:

The moment of inertia of the bar about the center of mass is

$I_r = 1888.80 \ kg m^2$

Explanation:

The free body diagram is shown on the second uploaded image

From the diagram we see that is

The mass of each segment is

$m_1 = \rho_1 l_1 w = 1 * 6 * 2 = 12$

$m_1 = \rho_2 l_2 w = 8 * 6 * 2 = 96$

$m_1 = \rho_2 l_2 w = 5 * 5 * 2 = 50$

The distance from the origin to the center of the segments i.e the center of masses for the individual segments

$x_2 = \frac{6}{2} + 6 = 9 m$

$x_3 = \frac{4}{2} + 12 = 14 m$

The resultant center of mass is mathematically evaluated as

$x_r = \frac{m_1 * x_1 + m_2 *x_2 + m_3 *x_3}{m_1 + m_2 + m_3}$

$= \frac{12 * 3 + 96 *9 + 50 *14}{12+ 96 + 50}$

$x_r = 10.13m$

The moment of Inertia of each segment of the bar is mathematically evaluated

$I_1 =\frac{m_1}{12}(l_1^2 + w^2)$ = $\frac{12}{12}(1^2 + 2^2)$

$I_1 = 4 \ kgm^2$

$I_2 =\frac{m_2}{12}(l_2^2 + w^2)$ = $\frac{96}{12}(6^2 + 2^2)$

$I_2 = 320 \ kgm^2$

$I_3 =\frac{m_3}{12}(l_3^2 + w^2)$ = $\frac{50}{12}(4^2 + 2^2)$

$I_2 = 83.334 \ kgm^2$

According to parallel axis theorem the moment of inertia about the center ( $x_r$ ) is mathematically evaluated as

$I_r = (I_1 + m_1 r_1^2) + (I_2 + m_2 r_2^2) +(I_3 + m_3 r_3^2)$

$I_r = (I_1 + m_1 |x_r - x_1|^2) + (I_2 + m_2 |x_r - x_2|^2) +(I_3 + m_3 |x_r - x_3|^2)$

$I_r = (4 + 12 |10.13 - 3|^2) + (320 + 96 |10.13 - 9|^2) +(83.334 + 50 |10.13 - 14|^2)$

$I_r = 1888.80 \ kg m^2$

6 0

3 years ago

A. For a 200g load acting vertically downwards at point B’, determine the axial load in members A’B’, B’C’, B’D’, C’D’ and C’E’.

Nonamiya [84]

Answer:

attached below

Explanation:

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3 years ago

Question 5 (20 pts) The rated current of a three-phase transmission line is 300 A. The currents flowing by the line are measured

prisoha [69]

Answer:

Check the explanation

Explanation:

Question 1.

The secondary current of 250/5 amps CT when 300 amps(rated current of transmission line ) flow in TL is

(5/250 ) X 300 = 6 amps

Question 2

The correct answer to this second question is yes, when Over current relay coil will operate and relay contacts gets close, if the pickup value( Ip) of relay is set as 6 amps in relay. ( because primary current of TL is 1.2 times of CT primary)

Question 3

Tap Block figure (Fig 1) is not available/uploaded in your question.

3 0

3 years ago

In the combination of resistors above, consider the 1.50 µΩ and 0.75 µΩ. How can you classify the connection between these two r

Airida [17]

Answer: they are connected in series.

Explanation:

3 0

3 years ago