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FinnZ [79.3K]
4 years ago
9

How does the energy used by a 100 W lightbulb compare with the energy used by a 40 W lightbulb in a given amount of time? The 10

0 W bulb uses 2.5 times more energy. The 40 W bulb uses 2.5 times more energy. The 100 W bulb uses 60 times more energy. The 40 W bulb uses 60 times more energy.
Physics
2 answers:
aleksandrvk [35]4 years ago
8 0

Answer:

The 100 W bulb uses 2.5 times more energy

Explanation:

just took the test

MatroZZZ [7]4 years ago
7 0
The correct answer is
<span>The 100 W bulb uses 2.5 times more energy

In fact, the power is the amount of energy used per unit of time. This means that the light bulb of 100 W uses 100 J in one second, and the light bulb of 40 W uses 40 J in one second, and if we compare the two numbers, we get
</span>\frac{100 J}{40 J}=2.5
<span>so, the 100 W light bulb uses 2.5 times more energy than the 40 W light bulb.</span>
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g initial angular velocity of 39.1 rad/s. It starts to slow down uniformly and comes to rest, making 76.8 revolutions during the
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Answer:

Approximately -1.58\; \rm rad \cdot s^{-2}.

Explanation:

This question suggests that the rotation of this object slows down "uniformly". Therefore, the angular acceleration of this object should be constant and smaller than zero.

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v^2 - u^2 = 2\, a\, x,

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The angular analogue of that equation will be:

(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta, where

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For this object:

  • \omega(\text{final}) = 0\; \rm rad\cdot s^{-1}, whereas
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The question is asking for an angular acceleration with the unit \rm rad \cdot s^{-1}. However, the angular displacement from the question is described with the number of revolutions. Convert that to radians:

\begin{aligned}\theta &= 76.8\; \rm \text{revolution} \\ &= 76.8\;\text{revolution} \times 2\pi\; \rm rad \cdot \text{revolution}^{-1} \\ &= 153.6\pi\; \rm rad\end{aligned}.

Rearrange the equation (\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta and solve for \alpha:

\begin{aligned}\alpha &= \frac{(\omega(\text{final}))^2 - (\omega(\text{initial}))^2}{2\, \theta} \\ &= \frac{-\left(39.1\; \rm rad \cdot s^{-1}\right)^2}{2\times 153.6\pi\; \rm rad} \approx -1.58\; \rm rad \cdot s^{-1}\end{aligned}.

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