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3 years ago

What is the intensity of a sound with a sound intensity level (SIL) 67 dB, in units of W/m^2?

Physics

1 answer:

vfiekz [6]3 years ago

8 0

Answer:

The intensity of sound (I) = 3.16 x 10⁻⁶ W/m²

Explanation:

We have expression for sound intensity level (SIL),

$L=10log_{10}\left ( \frac{I}{I_0}\right )$

Here we need to find the intensity of sound (I).

$L=10log_{10}\left ( \frac{I}{I_0}\right )\\\\log_{10}\left ( \frac{I}{I_0}\right )=0.1L\\\\\frac{I}{I_0}=10^{0.1L}\\\\I=I_010^{0.1L}$

Substituting

L = 67 dB and I₀ = 10⁻¹² W/m² in the equation

$I=I_010^{0.1L}=10^{-12}\times 10^{0.1\times 65}\\\\I_0=10^{-12}\times 10^{6.5}=10^{-5.5}=3.16\times 10^{-6}W/m^2$

The intensity of sound (I) = 3.16 x 10⁻⁶ W/m²

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2. A jack exerts a vertical force of 4.5 X 103

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Correct Question:-

A jack exerts a vertical force of 4.5 × 10³

newtons to raise a car 0.25 meter. How much

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Given :-

$\star \sf \small force = 4.5 \times {10}^{3} \: newton$

$\star \sf \small distance = 0.25 \: meter$

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To find:-

$\sf \star \: work = \: ?$

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Solution:-

we know :-

$\bf \dag \boxed{ \rm work = force \times distance}$

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$\dashrightarrow \sf work = force \times distance$

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$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times 0.5 \\$

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$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \frac{0 \cancel.5}{10} \\$

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$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \frac{5}{10} \\$

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$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \cancel \frac{5}{10} \\$

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$\dashrightarrow \sf work = \dfrac{4\cancel.5}{10} \times 1 {0}^{3} \times \dfrac{1}{2} \\$

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$\dashrightarrow \sf work = \dfrac{45}{10} \times 1 {0}^{3} \times \dfrac{1}{2} \\$

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$\dashrightarrow \sf work = \dfrac{45}{10 {}^{0} } \times 1 {0}^{3 - 1} \times \dfrac{1}{2} \\$

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$\dashrightarrow \sf work = \dfrac{45}{10 {}^{0} } \times 1 {0}^{2} \times \dfrac{1}{2} \\$

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$\dashrightarrow \sf work = \dfrac{45}{1} \times 1 {0}^{2} \times \dfrac{1}{2} \\$

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$\dashrightarrow \sf work = \dfrac{45 \times 10 \times \cancel{10}}{ \cancel2} \\$

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$\dashrightarrow \sf work = \dfrac{45 \times 10 \times 5}{ 1} \\$

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$\dashrightarrow \sf work =225 \times 10$

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$\dashrightarrow \bf work =\red{2250\: joule}$

5 0

3 years ago

An unbalanced force of 20N acts on a 4.0kg mass what is it's acceleration

tankabanditka [31]

Hi there!

According to Newton's second law:

∑F = m · a, where:

∑F = net force (N = kgm/s²)

m = mass (kg)

a = acceleration (m/s²)

Rearrange to solve for acceleration:

F/m = a

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3 years ago

Which of the following observations indicates that an object is falling at terminal velocity?

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Answer: D

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4 years ago

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