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2 years ago

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There may be installations where grounded conductors of different systems are installed within the same raceway, cable, or box.

In these cases, it's imperative that the system-grounded conductors remain
A. separated.
B. disconnected.
C. spliced.
D. buried.

1 answer:

Brrunno [24]2 years ago

7 0

It is imperative that the system-grounded conductors remain buried and is denoted as option D.

<h3>What is a Conductor?</h3>

This is referred as a substance which allows electricity and heat to flow through it and examples include metals.

These conductors have to be buried so as to prevent the risk of contact and electrocution of individuals thereby making it a safe method.

Read more about Conductor here brainly.com/question/492289

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A circular dartboard has a radius of 2 meters and a red circle in the center. Assume you hit the target at a random point. For w

Sati [7]

Answer:

1.549 m

Explanation:

Given:

The radius of the circular board, r = 2 m

The probability of hitting the red is given as 0.6

Now, this probability of hitting the red can be conclude as

0.6 = (Area of red)/ (Total area of the board)

Total area of the board = πr² = π × 2²

let the radius of the red area be R

thus, area of red circle, = πR²

on substituting the value of the area, we have

0.6 = (πR²)/ (π × 2²)

or

R² = 2.4

or

R = 1.549 m

Thus, the radius of the red circle is 1.549 m

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3 years ago

A steel ball is dropped to a wood table. What are three ways in which energy is wasted when the ball bounces?

hodyreva [135]

Answer:1. Inelastic deformation of wood and steel

2.thermal energy: hear from wood and steel

3.acoustic energy

Explanation:

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3 years ago

What is the value of the normal force if the coefficient of kinetic friction is 0.22 and the kinetic frictional force is 40 newt

ipn [44]

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3 years ago

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A crate is lifted vertically 1.5 m and then held at rest. The crate has weight 100 N (i.e., it is supported by an upward force o

bogdanovich [222]

Answer:

1) W = 150 J

Explanation:

Work (W) is defined as the product of force F by the distance (d)the body travels due to this force.

W= F*d Formula ( 1)

The work is positive (W+) if the force has the same direction of movement of the object.

The work is negative (W-) if the force has the opposite direction of the movement of the object.

The component of the force that performs work must be parallel to the displacement.

Work done to lift the floor box to its final position

We apply the formula (1)

W= F*d

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3 years ago

A block of mass m=2.20m=2.20 kg slides down a 30.0^{\circ}30.0

Xelga [282]

Answer:

$v_m \approx -4.38\; \rm m \cdot s^{-1}$ (moving toward the incline.)

$v_M \approx 4.02\; \rm m \cdot s^{-1}$ (moving away from the incline.)

(Assumption: $g = 9.81\; \rm m \cdot s^{-2}$ .)

Explanation:

If $g = 9.81\; \rm m \cdot s^{-2}$ , the potential energy of the block of $m = 2.20\; \rm kg$ would be $m \cdot g\cdot h = 2.20\; \rm kg \times 9.81\; \rm m \cdot s^{-2} \times 3.60\; \rm m \approx 77.695\; \rm J$ when it was at the top of the incline.

If friction is negligible, all these energies would be converted to kinetic energy when this block reaches the bottom of the incline. There shouldn't be any energy loss along the horizontal surface, either. Therefore, the kinetic energy of this $m = 2.20\; \rm kg\!$ block right before the collision would also be approximately $77.695\; \rm J$ .

Calculate the velocity of that $m = 2.20\; \rm kg$ based on its kinetic energy:

$\displaystyle v_m(\text{initial}) = \sqrt{\frac{2\times (\text{Kinetic Energy})}{m}} \approx \sqrt{\frac{2 \times 77.695\; \rm J}{2.20\; \rm kg}} \approx 8.4043\; \rm m \cdot s^{-1}}$ .

A collision is considered as an elastic collision if both momentum and kinetic energy are conserved.

Initial momentum of the two blocks:

$p_m = m \cdot v_m(\text{initial}) \approx 2.20\; \rm kg \times 8.4043\; \rm m \cdot s^{-1} \approx 18.489\; \rm kg \cdot m \cdot s^{-1}$ .

$p_M = M \cdot v_M(\text{initial}) \approx 2.20\; \rm kg \times 0\; \rm m \cdot s^{-1} \approx 0\; \rm kg \cdot m \cdot s^{-1}$ .

Sum of the momentum of each block right before the collision: approximately $18.489\; \rm kg \cdot m \cdot s^{-1}$ .

Sum of the momentum of each block right after the collision: $(m\cdot v_m + m \cdot v_M)$ .

For momentum to conserve in this collision, $v_m$ and $v_M$ should ensure that $m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1}$ .

Kinetic energy of the two blocks right before the collision: approximately $77.695\; \rm J$ and $0\; \rm J$ . Sum of these two values: approximately $77.695\; \rm J\!$ .

Sum of the energy of each block right after the collision:

$\displaystyle \left(\frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2\right)$ .

Similarly, for kinetic energy to conserve in this collision, $v_m$ and $v_M$ should ensure that $\displaystyle \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J$ .

Combine to obtain two equations about $v_m$ and $v_M$ (given that $m = 2.20\; \rm kg$ whereas $M = 7.00\; \rm kg$ .)

$\left\lbrace\begin{aligned}& m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1} \\ & \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J\end{aligned}\right.$ .

Solve for $v_m$ and $v_M$ (ignore the root where $v_M = 0$ .)

$\left\lbrace\begin{aligned}& v_m \approx -4.38\; \rm m\cdot s^{-1} \\ & v_M \approx 4.02\; \rm m \cdot s^{-1}\end{aligned}\right.$ .

The collision flipped the sign of the velocity of the $m = 2.20\; \rm kg$ block. In other words, this block is moving backwards towards the incline after the collision.

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2 years ago