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attashe74 [19]
4 years ago
15

A transformer connected to a 120--V ac line is to supply 12.0V (rms) to a portable electronic device. The load resistance in the

secondary is 4.70 ohms
1) What should the ratio of primary to secondary turns of the transformer be?

N1/N2 = _________

2) What rms current must the secondary supply?

I = _________ A

3) What average power is delivered to the load?

P = ________ W

4) What resistance connected directly across the source line (which has a voltage of 120?V ) would draw the same power as the transformer?

R = ________ ohms
Physics
1 answer:
Troyanec [42]4 years ago
3 0

We don't know anything about the structure or materials of the real transformer, so all we can do is assume that it's ideal, with no losses.

1). <em>N1/N2 =</em> 120v/12v = <em>10:1</em>

2). <em>I =</em> V/R = 12/4.7 = <em>2.55 Amperes</em>

3). <em>Power =</em> V²/R = (12)² / 4.7  =  <em>30.64 watts</em>

4). Power = V² / R

R = V² / power

R = (120)² / (30.64 watts)

<em>R =  470 ohms </em>

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Two people are talking at a distance of 3.0 m from where you are and you measure the sound intensity as 1.1 × 10-7 W/m2. Another
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Answer:I_2=0.618\times 10^{-7} W/m^2

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Read 2 more answers
A 98.0 N grocery cart is pushed 12.0 m along an aisle by a shopper who exerts a constant horizontal force of 40.0 N. If all fric
Digiron [165]

Answer:

9.8 m/s

Explanation:

The work done by the force pushing the cart is equal to the kinetic energy gained by the cart:

W=K_f -K_i

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K_i is the initial kinetic energy of the cart, which is zero because the cart starts from rest, so we can write:

W=K_f

But the work is equal to the product between the pushing force F and the displacement, so

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where m is the mass of the cart and v its final speed.

We can find the mass because we know the weight of the cart, 98.0 N:

m=\frac{F_g}{g}=\frac{98.0 N}{9.8 m/s^2}=10 kg

Therefore, we can now re-arrange eq.(1) to find the final speed of the cart:

v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(480 J)}{10 kg}}=9.8 m/s



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