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Flura [38]
4 years ago
14

(i) Two 110V lamps are connected in parallel. Their ratings are 150W and 200W.

Physics
1 answer:
LekaFEV [45]4 years ago
7 0

Answer:

37.7 Ω

Explanation:

The resistance in each lamp is:

P = IV = V²/R

R = V²/P

R₁ = (110 V)² / (150 W)

R₁ = 80.67 Ω

R₂ = (110 V)² / (200 W)

R₂ = 60.5 Ω

Their combined resistance in parallel is:

1/R = 1/R₁ + 1/R₂

1/R = 1/(80.67 Ω) + 1/(60.5 Ω)

R = 34.57 Ω

So the current that goes through them is:

V = IR

I = V/R

I = (110 V) / (34.57 Ω)

I = 3.182 A

We need the same current in the new circuit.

V = IR

230 V = (3.182 A) (R₃ + 34.57 Ω)

R₃ = 37.7 Ω

We can check our answer by finding the voltage drop across this resistor:

V = I R₃

V = (3.182 A) (37.7 Ω)

V = 120 V

230 V − 120 V = 110 V, the correct voltage for the lamps.

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A ball is rolled at a velocity of 25 m/sec. after 13.5 seconds, it comes to a stop . What is the acceleration of the ball?
nexus9112 [7]

Answer:

The answer to your question is a = -1.85 m/s² the acceleration is negative because it is coming to stop.

Explanation:

Data

vo = 25 m/s

t = 13.5 s

a= ?

vf = 0 m/s

Formula

vf = vo + at

solve for a

a = (vf - vo)/t

Substitution

a = (0 - 25) / 13.5

Simplification

a = -25/13.5

Result

a = -1.85 m/s²

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Answer:

1.7 m

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s_x=v_xt\\\Rightarrow s_x=4.47\times 0.38\\\Rightarrow s_x=1.7\ \text{m}

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