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Anarel [89]
2 years ago
15

Stu wanted to calculate the resistance of a light bulb connected to a 4.0 V battery, with a resulting current of 0.5 A. He used

the formula R = VI and obtained an answer of 2 Ω. Was Stu’s answer correct? How do you know?
Physics
2 answers:
Dimas [21]2 years ago
4 0
His answer was incorrect because according to ohm's law the formula used should have been R=V/I instead of multiplying and the answer should be 8ohms
Mashutka [201]2 years ago
4 0

Answer:

Sample Response: Stu's answer was incorrect. According to Ohm's law, R = V/I. This is the equation that should be used to obtain the correct answer, which is 8 ohms.

Explanation:

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The current supplied by a battery slowly decreases as the battery runs down. Suppose that the current as a function of time is:
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Answer: 8.1 x 10^24

Explanation:

I(t) = (0.6 A) e^(-t/6 hr)

I'll leave out units for neatness: I(t) = 0.6e^(-t/6)

If t is in seconds then since 1hr = 3600s: I(t) = 0.6e^(-t/(6 x 3600) ).

For neatness let k = 1/(6x3600) = 4.63x10^-5, then:

I(t) = 0.6e^(-kt)

Providing t is in seconds, total charge Q in coulombs is

Q= ∫ I(t).dt evaluated from t=0 to t=∞.

Q = ∫(0.6e^(-kt)

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= 0.6/k

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5 0
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Two convex thin lenses with focal lengths 10.0 cm and 20.0 cm are aligned on a common axis, running left to right, the 10-cm len
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Answer:

(c) +6.67

Explanation:

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\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{15}\\\Rightarrow \frac{1}{v}=\frac{1}{30}\\\Rightarrow v=30\ cm

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The final image will appear as +6.67 cm

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