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Paraphin [41]
3 years ago
8

A sound from a source has an intensity of 270 db when it is 1 m from the source. what is the intensity of the sound when it is 3

m from the source?
Physics
1 answer:
Lina20 [59]3 years ago
4 0
Sound intensity is inversely proportional to the square of the distance between the source and the receiver.
That is 
I = k/r^2
 where
 k = constant
 r = radius

When r=1,  the intensity is I₁ = k/1 = k
When r=3, the intensity I₂ = k/3² = k/9
Therefore
 I₂ = I₁ /9

In decibels,
I = 10 log₁₀(I/I₀)
where I₀ = reference intensity

When r=1,
10 log₁₀ (I₁/I₀) = 270

When r =3,
10 log₁₀ (I₂/I₀) = 10 log₁₀ [(I₂/I₁)*(I₁/I₀)]
                     = 10 log₁₀ [(1/9)*(I₁/I₀)]
                     = 10 log₁₀(1/9) + 270
                     = 260.5

Answer: 260.5 dB (nearest tenth)
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A 4.2-m-diameter merry-go-round is rotating freely with an angular velocity of 0.79 rad/s . Its total moment of inertia is 1790
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Answer:

w_2=0.467rad/s

Explanation:

Four people standing on the ground each of mass and usually this questions have to find the final angular velocity

m_t=4*70kg=280kg

The radius r=4.2/2=2.1m

Angular velocity  w_1=0.79rad/s

The moment of inertia total is I_t=1790 kg/m^2

Momento if inertia

I_1=m_t*r^2

I_1=280kg*(2.1m)^2=1234.8kg*m^2

Angular momentum

I_1*w_1=I_t*w_2

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w_2=\frac{I_1*w_1}{I_t}

w_2=\frac{1790kg*m^2*0.79rad/s}{3024.8kg*m^2}

w_2=0.467rad/s

8 0
3 years ago
Question C) needs to be answered, please help (physics)
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(a) Differentiate the position vector to get the velocity vector:

<em>r</em><em>(t)</em> = (3.00 m/s) <em>t</em> <em>i</em> - (4.00 m/s²) <em>t</em>² <em>j</em> + (2.00 m) <em>k</em>

<em>v</em><em>(t)</em> = d<em>r</em>/d<em>t</em> = (3.00 m/s) <em>i</em> - (8.00 m/s²) <em>t</em> <em>j</em>

<em></em>

(b) The velocity at <em>t</em> = 2.00 s is

<em>v</em> (2.00 s) = (3.00 m/s) <em>i</em> - (16.0 m/s) <em>j</em>

<em></em>

(c) Compute the electron's position at <em>t</em> = 2.00 s:

<em>r</em> (2.00 s) = (6.00 m) <em>i</em> - (16.0 m) <em>j</em> + (2.00 m) <em>k</em>

The electron's distance from the origin at <em>t</em> = 2.00 is the magnitude of this vector:

||<em>r</em> (2.00 s)|| = √((6.00 m)² + (-16.0 m)² + (2.00 m)²) = 2 √74 m ≈ 17.2 m

(d) In the <em>x</em>-<em>y</em> plane, the velocity vector at <em>t</em> = 2.00 s makes an angle <em>θ</em> with the positive <em>x</em>-axis such that

tan(<em>θ</em>) = (-16.0 m/s) / (3.00 m/s)   ==>   <em>θ</em> ≈ -79.4º

or an angle of about 360º + <em>θ</em> ≈ 281º in the counter-clockwise direction.

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