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3 years ago

Why cable color codes are important ? what if we use three phase cables without color

Physics

1 answer:

Degger [83]3 years ago

4 0

Answer:

Cable color codes are very important for example, If something is not wired properly, it can result in injuries, deaths, fires, and many other problems. This is why there are well-established wire color codes to ensure those working with an around this type of equipment can ensure everything is wired safely and effectively. If we used three phase cables without color code, this can lead too incorrect things that might be a hazard too get not fix and learning new color codes would be hard therefore we should stick too the same color codes for cables and it saves the hassle for needing too check every cable cord color then the original one signed too it.

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3 years ago

2. How do the phytochemicals present in various foods help us?<br>

vredina [299]

<h2>Answer:</h2>

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3 years ago

A 0.45-m metal rod moves 0.11 m in a direction that is perpendicular to a 0.80-T magnetic field in an elapsed time of 0.036 s. A

Gwar [14]

Answer:

1.1 V

Explanation:

L = 0.45 m

d = 0.11 m

B = 0.80 T

t = 0.036 s

Let e be the emf.

e = B v L

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4 years ago

Read 2 more answers

A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-

alexdok [17]

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, $q_1=-3\ nC$

It is placed at a distance of 9 cm at x axis

Charge, $q_2=+4\ nC$

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

$\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0$

Here,

$r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}$

So,

$\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}$

Squaring both sides,

$3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m$

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

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