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yanalaym [24]
3 years ago
9

The power grid is the system that transfers electricity to homes and businesses. A power plant produces electricity. The output

electricity is an input for the wires that lead to power sub stations. The power substations regulate the flow of electricity and send it as an output through different wires. Transformers adjust the electricity along the way until it becomes an input for a home. What is the primary type of input and output that is used in the power grid system
Physics
2 answers:
jok3333 [9.3K]3 years ago
8 0

Answer:

First of all, input electricity is just indicating the direction of that energy, in this case, it refers to the  electrity that is being received in a sub-system. On the other hand, output electricity is the energy that is being produced to another sub-system.

For example, in this case the power plant represents the production of energy, which offer output electricity because produces the energy for other sub-system. Then, those sub-systems receive this energy, that's why at this level it becomes input energy.

<em>Being said that, we can say that energy is the primary type of input and output, it's what is being used in the power g</em>rid.

sasho [114]3 years ago
6 0

Answer: energy

Explanation: So the input and output of the power grid system is energy.

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Answer the following questions for a mass that is hanging on a spring and oscillating up and down with simple harmonic motion. N
LiRa [457]

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1. equilibrium

2. bottom

3. bottom

4. nowhere

5. bottom

6. top & bottom

7. equilibrium

8. equilibrium

1. No

2. Yes

Explanation:

According to the following equation of motion for SHM:

x(t) = A\cos(\omega t + \phi)

where A is the amplitude, ω is the angular frequency, and ∅ is the phase angle.

Furthermore, the velocity and acceleration functions are as follows:

y(t) = -\omega A\sin(\omega t + \phi)\\a(t) = -\omega^2 A\cos(\omega t + \phi)

1. The acceleration is zero at the equilibrium. At the equilibrium, the net force on the object is zero. And according to Newton's Second Law, if the net force is zero, then the acceleration is zero as well.

2. The forces on the object in a vertical spring are the weight of the object and the spring force.

F = mg - kx

Since mg is constant along the motion, then the net force is maximum at the amplitude. For the special case in this question, the mass is always below the rest length of the spring. So the net force is maximum at the lower amplitude, because x is greater in magnitude at the lower amplitude.  According to Newton's Second Law, acceleration is proportional to the net force, hence the acceleration is at a maximum at the bottom.

3. As explained above, the magnitude of the net force is at a maximum at the lower amplitude, that is bottom.

4. The spring force is defined by Hooke's Law: F = -kx. Since the oscillation is small enough so that the mass is always below the rest length of the spring, then x is always greater than zero, hence nowhere in the motion will the spring force becomes zero.

5. As explained above, the force of gravity is constant and the spring force is proportional to the displacement, x. Therefore, the spring force is at a maximum at the lower amplitude, that is bottom.

6. The speed is zero when the mass is instantaneously at rest, that is the amplitude.

7. The net force on the mass is zero at the equilibrium.

8. The speed is at a maximum at the equilibrium.

1.  We will use the equation of motions given above. For simplicity, let's take ∅ = 0. At half its amplitude:

\frac{A}{2} = A\cos(\omega t)\\\frac{1}{2} = \cos(\omega t)\\\omega t = \pi / 3

Then the velocity at that point is

v(t) = -\omega A\sin(\pi /3) = -\omega A (0.866)

The maximum speed is where the acceleration is equal to zero:

0 = -\omega^2 A\cos(\omega t)\\\omega t = \pi / 2\\v_{max} = -\omega A\sin(\pi /2) = -\omega A

Comparing the maximum velocity to the velocity at A/2 yields that it is not half the maximum velocity:

-\omega A(0.866) \neq -\omega A

2. The maximum acceleration is at the amplitude.

A = A\cos(\omega t)\\\omega t = 2\pi\\a_{max} = -\omega^2 A\cos(2\pi) = -\omega^2 A

And the acceleration at A/2 is

\frac{A}{2} = A\cos(\omega t)\\\omega t = \pi / 3\\a(t) = -\omega^2 A\cos(\pi / 3) = -\omega^2 A (0.5)

Comparing these two results yields that the acceleration at half the amplitude is half the maximum acceleration.

5 0
3 years ago
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