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iogann1982 [59]
2 years ago
11

The period of a pendulum is the time it takes the pendulum to swing back and forth once. If the only dimensional quantities that

the period depends on are the acceleration of gravity, g, and the length of the pendulum, l, what combination of g and l must the period be proportional to
Physics
1 answer:
Vinvika [58]2 years ago
7 0

Explanation:

Let T is the period of a pendulum. The SI unit of time is seconds (s).

It depends on the acceleration of gravity, g, and the length of the pendulum, l.

The SI unit of acceleration of gravity, g and the length of the pendulum, l are m/s² and m respectively.

If we divide m and m/s², we left with s². If the square root of s² is taken, we get s only i.e. the SI unit of period of a pendulum.

So,

T\propto \sqrt{\dfrac{l}{g}}

Hence, this is the required solution.

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Answer:

<em>The purpose of an inclinded plane is to make easier to move objects to a certain height.</em>

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Answer:

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Explanation:

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Using the formula;

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E = \frac{\mu_o}{2 \pi}c In (\frac{b+a}{a}) I_o \omega cos \omega t

E_{rms} =   \frac { {\frac{\mu_o \ c}{2 \pi} In (\frac{b+a}{a}) I_o (2 \pi f)}}{\sqrt{2}}

Replacing our values into above equation; we have:

E_{rms} =   \frac { {\frac{4 \pi*10^{-7}*0.325}{2 \pi} In (\frac{72.2+80.2}{80.2}) *9.55 (2 \pi *359)}}{\sqrt{2}}

E_{rms} =   \frac {8.98909588*10^{-4} }{\sqrt{2}}

E_{rms} =   6.356*10^{-4} \ V

Then the P_{avg is calculated as:

P_{avg} = \frac{E^2}{R}

P_{avg} = \frac{(6.356*10^{-4})^2}{64.3}

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