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4 years ago

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Consider a transformer used to recharge rechargeable flashlight batteries, that has 500 turns in its primary coil, 4 turns in it

s secondary coil, and an input voltage of 120 V. What is the voltage output in volts, of the transformer used for to charge the batteries?

1 answer:

Ilia_Sergeevich [38]4 years ago

4 0

Answer:

Output Voltage = 0.96 volts

Explanation:

As we know by the equation of transformer we have

$\frac{V_1}{V_2} = \frac{N_1}{N_2}$

here we know that

$V_1$ = voltage of primary coil = 120 Volts

$V_2$ = voltage of secondary coil

$N_1$ = number of turns in primary coil = 500

$N_2$ = number of turns in secondary coil = 4

now we will have

$\frac{120}{V} = \frac{500}{4}$

$V = 0.96 Volts$

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Read 2 more answers

A force acts on a 9.90 kg mobile object that moves from an initial position of to a final position of in 5.40 s. Find (a) the wo

horrorfan [7]

Given that,

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Time =5.40 s

Suppose the force is (2.00i + 9.00j + 5.30k) N, initial position is (2.70i - 2.90j + 5.50k) m and final position is (-4.10i + 3.30j + 5.40k) m.

We need to calculate the displacement

Using formula of displacement

$s=r_{2}-r_{1}$

Where, $r_{1}$ = initial position

$r_{2}$ = final position

Put the value into the formula

$s= (-4.10i + 3.30j + 5.40k)-(2.70i - 2.90j + 5.50k)$

$s= -6.80i+6.20j-0.1k$

(a). We need to calculate the work done on the object

Using formula of work done

$W=F\cdot s$

Put the value into the formula

$W=(2.00i + 9.00j + 5.30k)\cdot (-6.80i+6.20j-0.1k)$

$W=-13.6+55.8-0.53$

$W=41.67\ J$

(b). We need to calculate the average power due to the force during that interval

Using formula of power

$P=\dfrac{W}{t}$

Where, P = power

W = work

t = time

Put the value into the formula

$P=\dfrac{41.67}{5.40}$

$P=7.71\ Watt$

(c). We need to calculate the angle between vectors

Using formula of angle

$\theta=\cos^{-1}(\dfrac{r_{1}r_{2}}{|r_{1}||r_{2}|})$

Put the value into the formula

$\theta=\cos^{-1}\dfrac{(-4.10i + 3.30j + 5.40k)\cdot(2.70i - 2.90j + 5.50k)}{7.54\times6.778})$

$\theta=79.7^{\circ}$

Hence, (a). The work done on the object by the force in the 5.40 s interval is 41.67 J.

(b). The average power due to the force during that interval is 7.71 Watt.

(c). The angle between vectors is 79.7°

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