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3 years ago

8

In a pith ball experiment, the two pith balls are at rest. The magnitude of the tension in each string is |T|=0.55N, and the ang

le between each string and a vertical line is θ=27.33∘. What are the values for the magnitudes of electrostatic force, Fq, and the gravitational force, Fg?

1 answer:

Len [333]3 years ago

5 0

Answer:

Explanation: Two pith ball will repel each other . they will remain balaced due to tension in the spring whose one component balances the weight and the other balances the repulsive force on each.

The gravitational force will be balanced by T cos 27.33 and the electrostatic repulsive force will be balanced by T sin27.33

So

Fg =T cos 27.33

= .55 X .888

= .49 N

Fq = T sin27.33

=.55 x .459

= .25 N.

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The speaker at the concert has the sound intensity level of 100 dB if we listen from the distance 5 m.How far from the speaker d

Mademuasel [1]

Answer:

$28.11m$ far from the speaker the intensity drops to 85 dB.

Explanation:

In the equation for the Decibel scale

$(1). \: \:\beta =10 log(\dfrac{I}{I_0})$

The ratio of the intensities can be written as

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$\dfrac{I}{I_0} = \dfrac{A_0}{A}.$

And since

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meaning

$\dfrac{I}{I_0} = \dfrac{r_0^2}{r^2}.$

Putting this into equation (1), we get:

$\boxed{ (2).\: \: \beta = 10log(\dfrac{r_0^2}{r^2})}$

Now, if the intensity is 100 dB when the distance is 5 meters, we have:

$100dB=10 log(\dfrac{r_02}{(5m)^2})$

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by taking both sides to the exponent:

$10^{10}= \dfrac{r_0^2}{25}$

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$\beta = 10log(\dfrac{25*10^{10}}{r^2})$

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$85 = 10log(\dfrac{25*10^{10}}{r^2})$

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take both sides to exponents and we get:

$10^{8.5} =10^{ log(\dfrac{25*10^{10}}{r^2})}$

$10^{8.5} =\dfrac{25*10^{10}}{r^2}$

$r^2 = \dfrac{25*10^{10}}{10^{8.5}}$

$\boxed{r = 28.11m}$

Thus, $28.11m$ far from the speaker the intensity drops to 85 dB.

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3 years ago