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Elis [28]
3 years ago
14

A bare 4 AWG copper conductor Installed horizontally near the bottom or vertically, and within that portion of a concrete foun d

ation or footing that is In direct contact with the earth, can be used as a grounding electrode when the conductor is at least __________ft in length.
Physics
1 answer:
neonofarm [45]3 years ago
4 0

Answer:

20 ft

Explanation:

This is the length where the direct contact could be used as ground electrode

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Which of the following is the FINAL step in a forecasting​ system?
Lelechka [254]

Answer:

gather the data needed to make the forecast

3 0
2 years ago
Find the voltage change when: a. An electric field does 12 J of work on a 0.0001-C charge. b. The same electric field does 24 J
kondaur [170]

Explanation:

Given that,

(a) Work done by the electric field is 12 J on a 0.0001 C of charge. The electric potential is defined as the work done per unit charged particles. It is given by :

V=\dfrac{W}{q}

V=\dfrac{12}{0.0001}

V=12\times 10^4\ Volt

(b) Similarly, same electric field does 24 J of work on a 0.0002-C charge. The electric potential difference is given by :

V=\dfrac{W}{q}

V=\dfrac{24}{0.0002}

V=12\times 10^4\ Volt

Therefore, this is the required solution.

7 0
3 years ago
If you were capable of converting mass to energy with 100%, efficiency, how much mass would you need to produce 3.5x10^12 Joules
Alexeev081 [22]

Answer:

a) 3.9 x 10⁻⁵ kg

Explanation:

The amount of mass required to produce the energy can be given by Einstein's formula:

E = mc^2\\\\m = \frac{E}{c^2}

where,

m = mass required = ?

E = Energy produced = 3.5 x 10¹² J

c = speed of light = 3 x 10⁸ m/s

Therefore,

m = \frac{3.5\ x\ 10^{12}\ J}{(3\ x\ 10^8\ m/s)^2} \\\\m = 3.9\ x\ 10^{-5}\ kg

Hence, the correct option is:

<u>a) 3.9 x 10⁻⁵ kg</u>

7 0
2 years ago
1. An astronaut in a spacesuit has a mass of 80 kilograms. What is the weight of this astronaut on the surface of the Moon where
Andrews [41]
Weight = 80 x (9.8/6) = .... N
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2 years ago
A 1-kg rock is suspended from the tip of a horizontal meterstick at the 0-cm mark so that the meterstick barely balances like a
tigry1 [53]

Explanation:

Given that,

Mass if the rock, m = 1 kg

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We need to find the mass of the meter stick. The force acting by the stone is

F = 1 × 9.8 = 9.8 N

Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.

9.8\times 12.5=W\times (50-12.5)\\\\W=\dfrac{9.8\times 12.5}{37.5}

W = 3.266 N

The mass of the meters stick is :

m=\dfrac{W}{g}\\\\m=\dfrac{3.266}{9.81}\\\\m=0.333\ kg

So, the mass of the meter stick is 0.333 kg.

5 0
3 years ago
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