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scoundrel [369]
3 years ago
14

A 12 V battery applies a potential difference to three resistors in series. If the potential at point P is 0.0 V, the potential

at point B is:
Physics
1 answer:
galina1969 [7]3 years ago
4 0

This question is incomplete, the complete question is on the image uploaded along this answer.

Answer:

the potential at point B is 5 V, Option d) is the correct answer

Explanation:

Given that;

from the image;

R = 3 + 4 + 5 = 12 Ω

so I = 12/12 = 1 A

Q = 0 + 12 = 12 V

now

VA - VQ = - I × 3 = -3 V

VA = VQ - 3 =  12 - 3 = 9 V

VB - VA = - I × 4Ω = - 4 V

⇒VB = VA - 4 = 9 V - 4 V = 5 V

Therefore the potential at point B is 5 V

Option d) is the correct answer

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A stone of mass m = 1.05 kg is released from a height of h = 2.1 m into a pool of water. At a time of t = 1.83 s after hitting t
mote1985 [20]

Answer:

Explanation:

ignoring air resistance, the kinetic energy at water impact will equal the potential energy converted

½mv² = mgh

v = √(2gh)

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FΔt = mΔv

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7 0
2 years ago
1. An object is traveling at a constant velocity of 30 m/s when it experiences a constant
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Answer:use

Explanation:google

7 0
3 years ago
Freight car A with a gross weight of 200,000 lbs is moving along the horizontal track in a switching yard at 4 mi/hr. Freight ca
zhenek [66]

Answer: a) 4.7 mi/hr.  b) 86,500 lbs. mi²/Hr²

Explanation:

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If the collision is fully inelastic, both masses continue coupled each other as a single mass, with a single speed.

So, we can write the following:

p₁ = p₂ ⇒m₁.v₁ + m₂.v₂ = (m₁ + m₂). vf

Replacing by the values, and solving for vf, we get:

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If the track is horizontal, this means that thre is no change in gravitational potential energy, so any loss of energy must be kinetic energy.

Before the collision, the total kinetic energy of the system was the following:

K₁ = 1/2 (m₁.v₁² + m₂.v₂²) = 3,400,000 lbs. mi² / hr²

After the collision, total kinetic energy is as follows:

K₂ = 1/2 ((m₁ + m₂) vf²) = 3,313,500 lbs. mi²/hr²

So we have an Energy loss, equal to the difference between initial kinetic energy and final kinetic energy, as follows:

DE = K₁ - K₂ = 86,500 lbs. mi² / hr²

This loss is due to the impact, and is represented by the work done by friction forces (internal) during the impact.

8 0
3 years ago
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