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NemiM [27]
1 year ago
12

A simple pendulum of length 2.5 m makes 5.0 complete swings in 16 s. What is the acceleration of gravity at the location?

Physics
1 answer:
Norma-Jean [14]1 year ago
8 0
<h2>The acceleration of gravity at the location is 9.64 m/s²</h2>

Explanation:

Length of pendulum = 2.5 m

Time taken for 5 swings = 16 seconds

Time taken for 1 swing = 3.2 seconds

Period of pendulum = 3.2 seconds.

We have equation for period of simple pendulum as

             T=2\pi \sqrt{\frac{l}{g}}

Where l is the length of pendulum and g is acceleration due to gravity.

Substituting

                 T=2\pi \sqrt{\frac{l}{g}}\\\\3.2=2\pi \sqrt{\frac{2.5}{g}}\\\\g=\frac{4\pi^2 \times 2.5}{3.2^2}\\\\g=9.64m/s^2

The acceleration of gravity at the location is 9.64 m/s²

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To solve the problem it is necessary to apply the concepts related to Conservation of linear Moment.

The expression that defines the linear momentum is expressed as

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v= velocity

According to our data we have to

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d=0.05m

A=13*10^6m^2

Volume (V) = A*d = (15*10^6)(0.03) = 3.9*10^5m^3

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From the given data we can calculate the volume of rain for 5 seconds

V' = \frac{V}{t}*\Delta t_{total}

Where,

\Delta t_{total} It is the period of time we want to calculate total rainfall, that is

V' = \frac{3.9*10^5}{10800}*5

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Through water density we can now calculate the mass that fell during the 5 seconds:

m' = V'*\rho

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Now applying the prevailing equation given we have to

P=m'v

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Therefore the momentum of the rain that falls in five seconds is 1.805*10^6 Kg.m/s

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Charges entering one meter of axon is 5.\times 10^{11} \times (+e)

Charges entering 0.100 mm of axon is 5.\times 10^{11} \times (+e) \times (0.1 \times 10^{-3}

substituting the value of +e=1.6\times 10^{-19} C in above equation, we get charge enter a 0.100 mm length of the axon is

q=5.\times 10^{11} \times1.6\times 10^{-19}  \times (0.1 \times 10^{-3}\\q=8.98\times 10^{-12} C

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2 years ago
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C

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