_____________ is an excessive current relative to normal operating current, but one that is confined to the normal conductive path provided by the conductors, circuit components, and loads of the distribution system. A(n) _________ is a current that flows outside the normal conducting path. One generally accepted definition of _______is when a phase or ungrounded conductor comes in contact with, or arcing current flows between, another phase conductor, neutral, or ground.

Answers

Answer 1
Answer:

Answer:

Overload current, short-circuit current and short circuit

Explanation:

Overload current is an excessive current relative to normal operating current, but one that is confined to the normal conductive path provided by the conductors, circuit components, and loads of the distribution system.

A short-circuit current is a current that flows outside the normal conducting path.

One generally accepted definition of short circuit is when a phase or ungrounded conductor comes in contact with, or arcing current flows between, another phase conductor, neutral, or ground.


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A cart has a momentum of 42 kg m/s to the right. Its mass is 6 kg. → What is its velocity? show your work

A water wave travels across a 42-meter wide pond in 7.0 seconds. The speed of the wave is _____. 292 m/s 49 m/s 6.0 m/s 0.17 m/s

Answers

           Speed  =  (distance covered) / (time to cover the distance).

The speed of anything that covers 42 meters in 7 seconds is

               (42 meters)  /  (7.0 seconds)

           =    (42 / 7.0)      (meters/second)

           =             6.0 m/s . 

Answer:

6.0 m/s

Explanation:

A water wave travels across a 42-meter wide pond in 7.0 seconds. The speed of the wave is _____.

292 m/s

49 m/s

6.0 m/s

0.17 m/s

The equation E = mc2 relates energy andA. force
B. work
C. gravity
D. mass

Answers

The equation E = mc² relates energy and mass and the correct option is option D.

The equation was formulated by Albert Einstein as part of his theory of special relativity. In this equation, "E" represents energy, "m" represents mass, and "c" represents the speed of light in a vacuum, which is approximately 3 x 10⁸ meters per second. The equation states that energy is equal to the mass of an object multiplied by the square of the speed of light.

This equation demonstrates the equivalence between mass and energy, suggesting that mass can be converted into energy and vice versa. It is a foundational equation in physics and has significant implications in understanding the relationship between matter and energy.

Thus, the ideal selection is option D.

Learn more about Energy, here:

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 E=mc2 is when energy is set equal to mass then it is multiplied by the velocity of light squared. The equation relates to energy and D.Mass

38. a 3.0 μc point charge is placed in an external uniform electric field that has a magnitude of 1.6 × 104 n/c. at what distance from the charge is the net electric field zero?

Answers

Answer:

To find the distance from the 3.0 μC point charge where the net electric field is zero in the presence of the external uniform electric field, you can use the principle that the electric fields due to the point charge and the external field will cancel each other at that point.

The electric field due to a point charge is given by Coulomb's law:

E_point_charge = k * (|q| / r^2),

where:

E_point_charge is the electric field due to the point charge.

k is Coulomb's constant (approximately 8.99 x 10^9 Nm^2/C^2).

|q| is the magnitude of the point charge (3.0 μC = 3.0 x 10^-6 C).

r is the distance from the point charge.

The external uniform electric field has a magnitude of 1.6 x 10^4 N/C. Let's denote this as E_external.

To find the point where the net electric field is zero, you want the magnitudes of the electric fields due to the point charge and the external field to be equal. So:

E_point_charge = E_external.

Substitute the expressions for both electric fields:

k * (|q| / r^2) = E_external.

Now, plug in the known values:

(8.99 x 10^9 Nm^2/C^2) * (3.0 x 10^-6 C / r^2) = 1.6 x 10^4 N/C.

Now, solve for r:

3.0 x 10^-6 C / r^2 = (1.6 x 10^4 N/C) / (8.99 x 10^9 Nm^2/C^2).

r^2 = (3.0 x 10^-6 C / (1.6 x 10^4 N/C)) * (8.99 x 10^9 Nm^2/C^2).

r^2 = (1.87 x 10^-11 m^2).

Take the square root of both sides to find r:

r ≈ 4.32 x 10^-6 m.

So, the net electric field is zero at a distance of approximately 4.32 x 10^-6 meters from the 3.0 μC point charge in the direction opposite to the external uniform electric field.

Explanation:

A hair dryer uses 72,000 joules of energy in 60 seconds what is the power of this hair dryer

Answers

Use work/time ratio.

 72,000 joules /60.0 seconds = 1,200 W.

How dense the medium is in the compression part of the wave, and how

Answers

How dense the medium is in the compression part of the wave and how rare the medium is in the rarefaction part of the wave is a measure of the longitudinal wave's amplitude.

What is the application for reflection of waves?

Answers

One big application is the manufacture and use of mirrors.