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2 years ago

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if the intensity of a person's voice is 4.6 x 10^-7 w/m^2 at a distance of 2.0 m, how much sound power does that person generate

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1 answer:

777dan777 [17]2 years ago

4 0

The sound power the person generated is $2.313 \times 10^{-5} \ W$ .

<h3>Area of the sound wave</h3>

The area of the sound wave is calculated as follows;

$A = 4\pi r^2\\\\A = 4 \pi \times (2)^2\\\\A = 50.272 \ m^2$

<h3>Power generated</h3>

The sound power the person generated is calculated as follows;

$P = I A\\\\P = 4.6\times 10^{-7} \ W/m^2 \ \ \times \ \ 50.272 \ m^2\\\\P = 2.313 \times 10^{-5} \ W$

Learn more about intensity of sound here: brainly.com/question/4431819

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To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,

$\theta = \frac{1.22 \lambda }{D}$

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The angle that relates the distance between the lights and the distance to the lamp is given by,

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For small angle, $sin\theta = \theta$

$sin \theta = \frac{d}{L}$

Here,

d = Distance between lights

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For small angle $sin \theta = \theta$

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3 years ago

A force of 1 N will cause a mass of 1 kg to have an acceleration of 1 m/s2. Therefore, a force of 7 N applied to a mass of 7 kg

Lana71 [14]

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2 years ago

Read 2 more answers

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NNADVOKAT [17]

I love science! if you need any more help let me know i cant guarantee i can help but i will try!

The correct answer is "Some substances must be dissolved in water before they can be used".

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3 years ago

An object dropped on Planet P falls 144 m in 6 seconds. What is the gravitational acceleration of Planet P ? Gravitational accel

Tju [1.3M]

Answer:

The gravitational acceleration of the planet is, g = 8 m/s²

Explanation:

Given data,

The distance the object falls, s = 144 m

The time taken by the object is, t = 6 s

Using the III equations of motion

S = ut + ½ gt²

∴ g = 2S/t²

Substituting the given values,

g = 2 x 144 /6²

= 8 m/s²

Hence, the gravitational acceleration of the planet is, g = 8 m/s²

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3 years ago

At one point in a pipeline, the water's speed is 3.57 m/s and the gauge pressure is 68.7 kPa. Find the gauge pressure at a secon

ArbitrLikvidat [17]

Answer:

The pressure at point 2 is $P_2 = 254.01 kPa$

Explanation:

From the question we are told that

The speed at point 1 is $v_1 = 3.57 \ m/s$

The gauge pressure at point 1 is $P_1 = 68.7kPa = 68.7*10^{3}\ Pa$

The density of water is $\rho = 1000 \ kg/m^3$

Let the height at point 1 be $h_1$ then the height at point two will be

$h_2 = h_1 - 18.5$

Let the diameter at point 1 be $d_1$ then the diameter at point two will be

$d_2 = 2 * d_1$

Now the continuity equation is mathematically represented as

$A_1 v_1 = A_2 v_2$

Here $A_1 , A_2$ are the area at point 1 and 2

Now given that the are is directly proportional to the square of the diameter [i.e $A= \frac{\pi d^2}{4}$ ]

which can represent as

$A \ \ \alpha \ \ d^2$

=> $A = c d^2$

where c is a constant

so $\frac{A_1}{d_1^2} = \frac{A_2}{d_2^2}$

=> $\frac{A_1}{d_1^2} = \frac{A_2}{4d_1^2}$

=> $A_2 = 4 A_1$

Now from the continuity equation

$A_1 v_1 = 4 A_1 v_2$

=> $v_2 = \frac{v_1}{4}$

=> $v_2 = \frac{3.57}{4}$

$v_2 = 0.893 \ m/s$

Generally the Bernoulli equation is mathematically represented as

$P_1 + \frac{1}{2} \rho v_1^2 + \rho * g * h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho * g * h_2$

So

$P_2 = \rho * g (h_1 -h_2 )+P_1 + \frac{1}{2} * \rho (v_1^2 -v_2 ^2 )$

=> $P_2 = \rho * g (h_1 -(h_1 -18.3) + P_1 + \frac{1}{2} * \rho (v_1^2 -v_2 ^2 )$

substituting values

$P_2 = 1000 * 9.8 (18.3) )+ 68.7*10^{3} + \frac{1}{2} * 1000 ((3.57)^2 -0.893 ^2 )$

$P_2 = 254.01 kPa$

8 0

2 years ago