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2 years ago

6

A book light has a 1.4 W, 4.8 V bulb that is powered by a transformer connected to a 120 V electric outlet. The secondary coil o

f the transformer has 20 turns of wire. How many turns does the primary coil have? What is the current in the primary coil?

1 answer:

VLD [36.1K]2 years ago

5 0

Answer:

Turns of the primary coil: 500

Current in the primary coil: Ip= 0.01168A

Explanation:

Considering an ideal transformer I can propose the following equations:

Vp×Ip=Vs×Is

Vp= primary voltaje

Ip= primary current

Vs= secondary voltaje

Is= secondary current

Np×Vs=Ns×Vp

Np= turns of primary coil

Ns= turns of secondary coil

From these equations I can clear the number of turns of the primary coil:

Np= (Ns×Vp)/Vp = (20×120V)/4.8V = 500 turns

To determine the current in the secondary coil I use the following equation:

Is= (1.4W)/4.8V = 0.292A

Therefore I can determine the current in the primary coil with the following equation:

Ip= (Vs×Is)/Vp = (4.8V×0.292A)/120V = 0.01168A

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3 years ago

Derive the equation of motion of the block of mass m1 in terms of its displacement x. The friction between the block and the sur

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Answer:

the equivalent mass : $m_e = m_1+m_2+\frac{I}{R^2}$

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Explanation:

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The velocity of the block to be = v

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The inertia of the cylinder about its center to be = I

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$v = \omega R$

$\omega =\frac{v}{R}$

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$K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I\omega^2$

By replacing $\omega$ with $\frac{v}{R}$ ; we have:

$K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I(\frac{v}{R})^2$

$K.E = \frac{1}{2}(m_1+ m_2+ \frac{I}{R} )v^2$ ------------------ equation (2)

Equating both equation (1) and (2); we have:

$m_e = m_1+m_2+\frac{I}{R^2}$

Therefore, the equivalent mass : $m_e = m_1+m_2+\frac{I}{R^2}$ which is read as;

The equivalent mass is equal to the mass of the block plus the mass of the cylinder plus the inertia by the square of the radius.

The expression for the force acting on equivalent mass due to the block is as follows:

$f_{block }=m_1gsin \beta$

Also; The expression for the force acting on equivalent mass due to the cylinder is as follows:

$f_{cylinder} = m_2gsin \phi$

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$m_e \bar x = f_{cylinder}-f_{block}$

which can be written as follows from the previous derivations

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<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of the isotopes each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

We are given:

For isotope 1:

Mass of isotope 1 = 19.99 amu

Percentage abundance of isotope 1 = 90.92 %

Fractional abundance of isotope 1 = 0.9092

For isotope 2:

Mass of isotope 2 = 20.99 amu

Percentage abundance of isotope 2 = 0.26%

Fractional abundance of isotope 2 = 0.0026

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Mass of isotope 3 = 21.99 amu

Percentage abundance of isotope 3 = 8.82%

Fractional abundance of isotope 3 = 0.0882

Putting values in equation 1, we get:

$\text{Average atomic mass}=[(19.99\times 0.9092)+(20.99\times 0.0026)+(21.99\times 0.0882)]$

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