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ki77a [65]
2 years ago
8

A 1500-W electric heater is plugged into the outlet of a 120-V circuit that has a 20-A circuit breaker. You plug an electric hai

r dryer into the same outlet. The hair dryer has power settings of 600 W, 900 W, 1200 W, and 1500 W. You start with the hair dryer on the 600-W setting and increase the power setting until the circuit breaker trips.
What power setting caused the breaker to trip?A) 600 WB) 900 WC) 1200 WD) 1500 W
Physics
1 answer:
Korolek [52]2 years ago
8 0

Answer:

b) 900 W

Explanation:

The breaker trips when the current is equal to 20 A. The power (P) is the ddp (V) multiplied by the current. So, for the electric heater, the current is:

P = V*i

1500 = 120*i

i = 12.5 A

So, to become in 20 A, it's needed 7.5 A, which must come from the hairdryer. Its power must be:

P = 120*7.5

P = 900 W

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HURRY! PLEASE HELP!!!!<br><br><br> 3. What methods are you using to test this (or each) hypothesis?
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2 years ago
A 2. 0 μf and a 4. 0 μf capacitor are connected in series across an 8. 0-v dc source. what is the charge on the 2. 0 μf capacito
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voltage across 2.0μf capacitor is 5.32v

Given:

C1=2.0μf

C2=4.0μf

since two capacitors are in series there equivalent capacitance will be

[tex] \frac{1}{c} = \frac{1}{c1} + \frac{1}{c2} [/tex]

c =  \frac{c1 \times c2}{c1 + c2}

=  \frac{2 \times 4}{2 + 4}

=1.33μf

As the capacitance of a capacitor is equal to the ratio of the stored charge to the potential difference across its plates, giving: C = Q/V, thus V = Q/C as Q is constant across all series connected capacitors, therefore the individual voltage drops across each capacitor is determined by its its capacitance value.

Q=CV

given,V=8v

= 1.33 \times 10 {}^{ - 6}  \times 8

= 10.64 \times 10 {}^{ - 6}

charge on 2.0μf capacitor is

\frac{Qeq}{2 \times 10 {}^{ - 6} }

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learn more about series capacitance from here: brainly.com/question/28166078

#SPJ4

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1 year ago
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