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tatuchka [14]
2 years ago
15

How does the circuit change when the wire is added? a closed circuit occurs and makes all bulbs turn off. an open circuit occurs

and makes all bulbs turn off. a short circuit occurs and makes bulbs 3 and 4 turn off but keeps bulbs 1 and 2 lit. a short circuit occurs and makes bulbs 1 and 2 turn off but keeps bulbs 3 and 4 lit.
Physics
1 answer:
Vitek1552 [10]2 years ago
3 0

The circuit change when a wire is added is, an open circuit occurs and makes all bulbs turn off.

<h3>What is a closed circuit?</h3>

A closed circuit is a type of circuit connection in which the wire connection is complete and current flow occurs, turning the light bulbs on in the process.

<h3>What is an open circuit?</h3>

An open circuit is a type of circuit connection in which the wire connection is incomplete and current cannot flow, turning off the light bulbs.

Thus, the circuit change when the wire is added is, an open circuit occurs and makes all bulbs turn off.

Learn more about open circuit here: brainly.com/question/20351910

#SPJ4

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Radio waves: The electromagnetic spectrum's longest wavelengths, which are found in radio waves, are normally found at frequencies of 300 gigahertz and below.

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if two point charges are separated by 1.5 cm and have charge values of 2.0 and -4.0, respectively, what is the value of the mutu
RUDIKE [14]

Complete question:

if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

The mutual force between the two point charges is 319.64 N

Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

value of the charges, q₁ and q₂ = 2 μC and - μ4 C

Apply Coulomb's law;

F = \frac{k|q_1||q_2|}{r^2}

where;

F is the force of attraction between the two charges

|q₁| and |q₂| are the magnitude of the two charges

r is the distance between the two charges

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N

Therefore, the mutual force between the two point charges is 319.64 N

4 0
3 years ago
REMARKS The speed found in part (a) is the same as if the woman fell vertically through a distance of 21.9 m. The result of part
sasho [114]

Answer:

Yes, if the system has friction, the final result is affected by the loss of energy.

Explanation:

The result that you are showing is the conservation of mechanical energy between two points in the upper one, the energy is only potential and the lower one is only kinetic.

In the case of some type of friction, the change in energy between the same points is equal to the work of the friction forces

    W_{fr} = ΔEm

    W_{fr} = Em_{f} -Em₀

As we can see now there is another quantity and for which the final energy is lower and therefore the final speed would be less than what you found in the case without friction.

    Em_{f} =W_{fr} + Em₀

 

Remember that the work of the rubbing force is negative, let's write the work of the rubbing force explicitly, to make it clearer

    ½ m v² = -fr d + mgh

    v = √(-fr d 2/m + 2 gh)

    v = √ (2gh - 2fr d/m)

Now it is clear that there is a decrease in the final body speed.

Consequently, if the system has friction, the final result is affected by the loss of energy.

5 0
2 years ago
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