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3 years ago

If a speaker gives a sound intensity of 10−6w/m2 at a certain point, what is the sound intensity level β at that point?

1 answer:

5 0

Start with basic concept that you have the same amount of watts it's just how far away you are dictates the watts/meter squared (per area). As the distance from the object is increased, the area increases as distance squared (think of a sphere of increasing radius, as you increase the radius, the surface area increases as r^2, double radius, you get 4 times area, triple radius, 9 times area, and so forth). so safe from acoustical effects (reflections off walls, etc) as you increase the distance from a speaker, the sound will decrease by a factor of the distance squared, as before 2x distance =1/4 power, 3x distance =1/9 power.

so the only thing left out of your questions is what distance from speaker is the level at 10^-6 W/m^2 and what distance is point B from the speaker, and apply the 1/distance^2 principle. 

The power at point B (PB) is the following: 

power at given power level (we will call this point A or PA). 

PB=PA*(DA/DB)^2 

Where PB is power level at PB (in W/m^2) 
PA=power at point A or 10^-6 W/m^2 
DA=distance from speaker to point A 
DB=distance from speaker to point B.

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4 years ago

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A particle moves in a straight line and has acceleration given by a(t) = 12t + 10. Its initial velocity is v(0) = −5 cm/s and it

soldi70 [24.7K]

Answer:

The position function is $s_{t}=2t^3+5t^2-5t+9$ .

Explanation:

Given that,

Acceleration $a =12t+10$

Initial velocity $v_{0} = -5\ cm/s$

Initial displacement $s_{0}=9\ cm$

We know that,

The acceleration is the rate of change of velocity of the particle.

$a = \dfrac{dv}{dt}$

The velocity is the rate of change of position of the particle

$v=\dfrac{dx}{dt}$

We need to calculate the the position

The acceleration is

$a_{t} = 12t+10$

$\dfrac{dv}{dt} = 12t+10$

$a_{t}=dv=(12t+10)dt$

On integration both side

$\int{dv}=\int{(12t+10)}dt$

$v_{t}=6t^2+10t+C$

At t = 0

$v_{0}=0+0+C$

$C=-5$

Now, On integration again both side

$v_{t}=\int{ds_{t}}=\int{(6t^2+10t-5)}dt$

$s_{t}=2t^{3}+5t^2-5t+C$

At t = 0

$s_{0}=0+0+0+C$

$C=9$

$s_{t}=2t^3+5t^2-5t+9$

Hence, The position function is $s_{t}=2t^3+5t^2-5t+9$ .

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3 years ago

When checking for noncondensables inside a recovery cylinder why should the technician allow the temperature of the cylinder to

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3 years ago

Will give correct answer brainliest

Katarina [22]

Adding my answer for points thank you everyone for this

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3 years ago

A 1.00 -kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Fig. P8.62a). The object has

LekaFEV [45]

The distance D where the object comes to rest is 1.08.m.

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The separation of one thing from another in space; the distance or separation in space between two objects, points, lines, etc.; remoteness. The distance of seven miles cannot be accomplished in one hour of walking.
Learn how to use the Pythagorean theorem to get the separation between two points using the distance formula. The Pythagorean theorem can be rewritten as d==(((x 2-x 1)2+(y 2-y 1)2)
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$W_{total} =$ ΔKE ⇒ -0.25*1*9.8*D = 0-1/2*1* $2.3^{2}$

⇒ $D=\frac{0.5*2.3^{2} }{2.45}$

⇒1.08.m

To learn more about distance, refer to:

brainly.com/question/4998732

#SPJ4

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2 years ago