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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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A high-speed flywheel in a motor is spinning at 450 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and
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Answer:

A) \omega_f=17.503\ rad.s^{-1}

B) t=55.6822\ s

C) \theta=1312\ rad

Explanation:

Given:

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  • diameter of flywheel, d=0.72\ m
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  • duration for which the power is off, t_0=35\ s
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<u>Using equation of motion:</u>

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

360\pi=15\pi\times 35+\frac{1}{2} \times \alpha\times35^2

\alpha=-0.8463\ rad.s^{-2}

Negative sign denotes deceleration.

A)

Now using the equation:

\omega_f=\omega_i+\alpha.t

\omega_f=15\pi-0.8463\times 35

\omega_f=17.503\ rad.s^{-1} is the angular velocity of the flywheel when the power comes back.

B)

Here:

\omega_f=0\ rad.s^{-1}

Now using the equation:

\omega_f=\omega_i+\alpha.t

0=15\pi-0.8463\times t

t=55.6822\ s is the time after which the flywheel stops.

C)

Using the equation of motion:

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

\theta=15\pi\times 55.68225-0.5\times 0.8463\times 55.68225^2

\theta=1312\ rad revolutions are made before stopping.

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mina [271]
1. Find the force of friction between the sports car and the station wagon stuck together and the road. The total mass m = 1928kg + 1041kg = 2969kg. The only force in the x-direction is friction: F = μ*N = μ * m * g 
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x = 0.5*a*t² 
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m₁*v₁ + m₂*v₂ = (m₁+m₂) * v = m₁*v₁ 
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v₁ = 33.9 m/s


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