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Vesnalui [34]
3 years ago
15

The charger for your electronic devices is a transformer. Suppose a 60 HzHz outlet voltage of 120 VV needs to be reduced to a de

vice voltage of 3.0 VV. The side of the transformer attached to the electronic device has 60 turns of wire.How many turns are on the side that plugs into the outlet?
Physics
1 answer:
photoshop1234 [79]3 years ago
4 0

Answer:

2400 turns

Explanation:

Applying

Vs/Vp = Ns/Np.................. Equation 1

Where Vs = Secondary voltage of the transformer, Vp = Primary voltage of the transformer, Ns = number of turns in the secondary side, Np = Number of turns in the primary side.

First we make Np the subject of the equation

Np = NsVp/Vs................. Equation 2

Given: Ns = 60 turns, Vp = 120 v Vs = 3.0 V

Substitute into equation 2

Np = 60(120)/3

Np = 2400 turns

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Answer:

Explanation:

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So, the conservation of energy in elastic collision leads to following equation:

M_{1} u_{1} +M_{2} u_{2}=M_{1}  v_{1}+M_{2}  v_{2}

Since, the momentum is conserved ,the kinetic energy will also be conserved in elastic collision. So

M_{1} u_{1} ^{2}+M_{2} u_{2} ^{2}=M_{1}v_{1} ^{2}+  M_{2}v_{2} ^{2}

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M_{2} u_{2}=M_{1}  v_{1}+M_{2}  v_{2}

and

M_{2} u_{2} ^{2}=M_{1}v_{1} ^{2}+  M_{2}v_{2} ^{2}

So, on solving all the above equation, we get an equation for velocity and that is

\frac{2M_{2}u_{2} }{(M_{1}+M_{2}  }=final velocity of ball with mass 2.5 kg

v = \frac{2(5*3.5)}{2.5+5}=4.67 m/s

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Where;

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Whereby the magnetic force acting on the charge particle = The centripetal force, we have;

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(1/2) × r × F_c = (1/2) × r × m·v²/r = (1/2)·m·v² = K.E.

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