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2 years ago

14

If it takes 200 joules of energy to lift a bucket of water 3 meters in 2 seconds, how much power would be required to do the sam

e amount of work in 5 seconds?

1 answer:

Harlamova29_29 [7]2 years ago

3 0

200 joules of work energy are involved. That's all we need to know to answer the question. Once we know that 200 joules of work energy are involved, we don't care what was lifted, or how far, or how long it took, or how many people worked on it, or how much they were paid, or what was the distribution of their gender identities, or the ethnic diversity among the team. or what day each of them celebrates as their sabbath. Any other information besides the 200 joules is only there to distract us, and see whether we're paying attention.

Power = (work or energy) / (time to do the work or move the energy)

Power = (200 joules) / (5 seconds)

<em>Power = 40 watts</em>

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The fundamental frequency of the tube is 0.240 m long, by taking air temperature to be $37^o$ C is 367.42 Hz.

A standing wave is basically a superposition of two waves propagating opposite to each other having equal amplitude. This is the propagation in a tube.

The fundamental frequency in the tube is given by

$f=\frac{v_T}{4L}$

where, $v_T=v\sqrt{\frac{T}{273} }$

Since, T=37+273 K = 310 K

v = 331 m/s

$\therefore v_T=331\sqrt{\frac{310}{273} } = 352.72 \ m/s$

Using this, we get:

$f=\frac{352.72}{4(0.240)} \\f=367.42 \ Hz$

Hence, the fundamental frequency is 367.42 Hz.

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11 months ago

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Answer:

a

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b

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Stranded Wire $R_1 = 0.0189 \ \Omega$

Explanation:

Considering the first question

From the question we are told that

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Considering the first wire

The cross-sectional area of the first wire is

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The cross-sectional area of the second wire is

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Generally the current is

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=> $I_2 = 1750 * 5.5899 *10^{-6}$

=> $I_2 = 0.00978 \ A$

Considering question two

From the question we are told that

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Generally the resistance of the first wire is mathematically represented as

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=> $R = \frac{ 1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }$

=> $R = 0.0149 \ \Omega$

Generally the resistance of the first wire is mathematically represented as

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