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Vesna [10]
3 years ago
6

The intensity level of a "Super-Silent" power lawn mower at a distance of 1.0 m is 100 dB. You wake up one morning to find that

four of your neighbors are all mowing their lawns using identical "Super-Silent" mowers. When they are each 20 m from your open bedroom window, what is the intensity level of the sound in your bedroom? You can neglect any absorption, reflection, or interference of the sound. The lowest detectable intensity is 1.0 × 10-12 W/m2. The intensity level of a "Super-Silent" power lawn mower at a distance of 1.0 m is 100 dB. You wake up one morning to find that four of your neighbors are all mowing their lawns using identical "Super-Silent" mowers. When they are each 20 m from your open bedroom window, what is the intensity level of the sound in your bedroom? You can neglect any absorption, reflection, or interference of the sound. The lowest detectable intensity is 1.0 × 10-12 W/m2. 40 dB 400 dB 104 dB 50 dB 80 dB
Physics
1 answer:
ANEK [815]3 years ago
4 0

Answer:

The 80 dB option is correct.

Explanation:

The sound intensity level is converted to sound intensity with the formula

[D] = 10 log (I/I₀)

I₀ = 10⁻¹² W/m²

[D] = 100 dB

100 = 10 log (I/I₀)

Log (I/I₀) = 10

(I/I₀) = 10¹⁰

I = I₀ × 10¹⁰ = 10⁻¹² × 10¹⁰ = 10⁻² W/m²

But sound intensity varies inversely as the square of the distance of the sound.

I ∝ (1/d²)

I = k/d²

At d = 1 m away, I = 10⁻² W/m²

0.01 = k/1

k = 0.01 W

At d = 20 m, we solve for I

I = 0.01/20² = 2.5 × 10⁻⁵ W/m²

4 people mowing their lawns at the same time,

I = 4 × 2.5 × 10⁻⁵ = 10⁻⁴ W/m⁻²

The sound intensity level in decibels is given as

[D] = 10 log (I/I₀)

[D] = 10 log (10⁻⁴/10⁻¹²)

[D] = 10 log (10⁸)

[D] = 10 × 8 = 80dB

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Two engineering students, John with a weight of 96 kg and Mary with a weight of 48 kg, are 30 m apart. Suppose each has a 0.04%
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Answer:

6.8370869499\times 10^{20}\ N

Explanation:

N_A = Avogadro's number = 6.022\times 10^{23}

e = Charge of electron = 1.6\times 10^{-19}\ C

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Z = Atomic number of water = 18

M = Molar mass of water = 0.018 kg/mol

m = Mass of person

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q=imbalance\times n\times e

Total number of protons and electrons in each sphere

n=\dfrac{mN_AZe}{M}

q=imbalance\times \dfrac{mN_AZe}{M}

q_1=0.0004\times \dfrac{96\times 6.022\times 10^{23}\times 18\times 1.6\times 10^{-19}}{0.018}\\\Rightarrow q_1=3699916.8\ C

q_2=0.0004\times \dfrac{48\times 6.022\times 10^{23}\times 18\times 1.6\times 10^{-19}}{0.018}\\\Rightarrow q_1=1849958.4\ C

Electrical force is given by

F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^{9}\times 3699916.8\times 1849958.4}{30^2}\\\Rightarrow F=6.8370869499\times 10^{20}\ N

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a painting in an art gallery has height h and is hung so that its lower edge is a distance d above the eye of an observer. How f
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Solution:

With reference to Fig. 1

Let 'x' be the distance from the wall

Then for \DeltaDAC:

tan\theta = \frac{d}{x}

⇒ \theta = tan^{-1} \frac{d}{x}

Now for the \DeltaBAC:

tan\theta = \frac{d + h}{x}

⇒ \theta = tan^{-1} \frac{d + h}{x}

Now, differentiating w.r.t x:

\frac{d\theta }{dx} = \frac{d}{dx}[tan^{-1} \frac{d + h}{x} -  tan^{-1} \frac{d}{x}]

For maximum angle, \frac{d\theta }{dx} = 0

Now,

0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} -  tan^{-1} \frac{d}{x}][/tex]

0 = \frac{-(d + h)}{(d + h)^{2} + x^{2}} -\frac{-d}{x^{2} + d^{2}}

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After solving the above eqn, we get

x = \sqrt{\frac{d}{d + h}}

The observer should stand at a distance equal to x = \sqrt{\frac{d}{d + h}}

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