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2 years ago

When a new path of lesser resistance is made for an existing circuit a(n) _____________ circuit occurs.

Physics

1 answer:

Margaret [11]2 years ago

5 0

When a new path of lesser resistance is made for an existing circuit a(n) short circuit occurs.

<h3>What is short circuit?</h3>

An electrical circuit short circuit is when two nodes that are supposed to be connected at different voltages make an improper connection. This leads to an electric current that can damage circuits, cause overheating, fire, or explosions, and is only constrained by the network's remaining nodes' equivalent Thevenin resistance. While short circuits are typically the result of a failure, they can occasionally be brought on purpose, such as when voltage-sensing crowbar circuit protectors are being installed.

An electrical connection that requires two nodes to have the same voltage is known as a short circuit in circuit analysis. Since there is no resistance and hence no voltage drop across the link in a "perfect" short circuit, there is no short circuit.

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The velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

Explanation:

Given;

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mass of the wooden block, m₂ = 2.43 kg

initial velocity of the wooden block, u₂ = 0

height reached by the bullet-block system after collision = 0.295 cm = 0.00295 m

let the initial velocity of the bullet on leaving the gun's barrel = v₁

let final velocity of the bullet-wooden block system after collision = v₂

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The kinetic energy of the bullet-block system after collision;

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K.E = ¹/₂ (2.4325)v₂²

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¹/₂ (2.4325)v₂² = 0.07032

1.21625 v₂² = 0.07032

v₂² = 0.07032 / 1.21625

v₂² = 0.0578

v₂ = √0.0578

v₂ = 0.24 m/s

Substitute v₂ in equation (1), to obtain the initial velocity of the bullet;

0.00247v₁ = 2.4325v₂

0.00247v₁ = 2.4325 (0.24)

0.00247v₁ = 0.5838

v₁ = 0.5838 / 0.00247

v₁ = 236.36 m/s

Therefore, the velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

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