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stiks02 [169]
4 years ago
13

A wire delivers 12.0 C of charge in 4.0 s. What is the current in the wire? 3.0 A 8.0 A 16 A 48 A

Physics
2 answers:
poizon [28]4 years ago
6 0

Q = Charge delivered by the wire = 12.0 C

t = time taken for the charge to be delivered = 4.0 s

i = current flowing in the wire due to the flow of charge  = ?

Current is defined as the time rate at which charge pass through a particular point.

Current in the wire is given as

i = Q/ t

inserting the values

i = (12.0) /(4.0)

i = 3 A

hence the correct choice is 3.0 A

const2013 [10]4 years ago
5 0
3.0 A i got it off Quizlet and there usually always right lol can't submit tho my answers to short.... Dot dot dot
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Answer:

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Explanation:

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Nesterboy [21]

Answer:

<em>The velocity after the collision is 2.82 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

It states the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is  

P=mv.  

If we have a system of two bodies, then the total momentum is the sum of the individual momentums:

P=m_1v_1+m_2v_2

If a collision occurs and the velocities change to v', the final momentum is:

P'=m_1v'_1+m_2v'_2

Since the total momentum is conserved, then:

P = P'

Or, equivalently:

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

If both masses stick together after the collision at a common speed v', then:

m_1v_1+m_2v_2=(m_1+m_2)v'

The common velocity after this situation is:

\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}

There is an m1=3.91 kg car moving at v1=5.7 m/s that collides with an m2=4 kg cart that was at rest v2=0.

After the collision, both cars stick together. Let's compute the common speed after that:

\displaystyle v'=\frac{3.91*5.7+4*0}{3.91+4}

\displaystyle v'=\frac{22.287}{7.91}

\boxed{v' = 2.82\ m/s}

The velocity after the collision is 2.82 m/s

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olga2289 [7]

Answer:

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