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dolphi86 [110]
3 years ago
12

Architects have to consider how sound travels when designing large buildings such as theaters, auditoriums, or museums. Sound in

tensity I is inversely proportional to the square of the distance from the sound source d. When the distance is 2, the sound intensity is 4. What is the intensity when the distance is 20
Physics
1 answer:
serious [3.7K]3 years ago
5 0

Answer:

  I₂ = 0.04 W / m²

Explanation:

Sound intensity is the power emitted between the unit area

          I = W / A

         W = I A

sound is a wave that travels in space whereby its energy is distributed on the surface of a sphere

         A = 4π r²

we substitute

         W = I (4π r²)

the emission power is constant, so the intensity at two different points is

           W = I₁ 4π r₁² = I₂ 4π r₂²

so the equation is

            I₁ r₁² = I₂ r₂²

In this case the units are not shown in the exercise, suppose that all units are in the SI system

           I₂ = I_1 \ \frac{r_1^2}{r_2^2}

let's calculate

           I₂ = 4 4 \  \frac{2^2}{20^2}

           I₂ = 0.04 W / m²

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For a circuit shown in the figure, all quantities are accurate in 3 significant figures. What is the power dissipated in 2-ohm r
sergey [27]
Base on your diagram it is a Series-Parallel type of connections right? So based on my calculation and the rule in computing the resistance and voltage of a parallel and series. The total resistance is 1.25 and the total voltage is 2.04 V
7 0
4 years ago
With what minimum speed must you toss a 130 g ball straight up to just touch the 14-m-high roof of the gymnasium if you release
Dmitry [639]

THIS IS THE COMPLETE QUESTION BELOW

With what minimum speed must you toss a 130 g ball straight up to just touch the 14-m-high roof of the gymnasium if you release the ball 1.1 m above the ground

And what speed does the ball hit the ground? Solve this problem using energy.

Answer

a)minimum speed must you toss a 130 g is 15.9090m/s

b)speed the ball hit the ground is 16.57m/s

Explanation:

a)We know that For any closed/isolated system, the total energy is CONSERVED.

K.E. lost by the ball=The change in P.E of the ball at 1.1m above the ground as well as the P.E. of the ball at 14 m-high roof

This statement can be expressed as the expression below from K.E and P.E energy formula

P.E. = mgh

K.E. = (1/2)mv^2

Therefore,

(mgh1 - mgh2)=(1/2)mv^2

Where h2=the ball height above the ground=1.1m

h1=ball height at roof of the gymnasium= 14m

Then if we substitute we have

[(10) x (0.14) x (9.81)] - [(1.1) x (0.14) x (9.81)] = (1/2)(0.14)(v^2)

16.45137=0.065V^2

V=15.9090m/s

minimum speed must you toss a 130 g is 15.9090m/s

b)To calculate the speed the ball hit the ground?

This is the highest point (14m-high roof),and the type of the energy the ball possesses is Po.tential energy only.

At the lowest point (ground), the energy the ball possesses is K.E. only.

P.E at 10m-high roof = K.E. at ground.

(14) x (0.13) x (9.81) = (1/2) x (0.13) x v^2

17.8542= 0.065V^2

V= 16.57

Therefore,And speed the ball hit the ground is 16.57m/s

6 0
4 years ago
Need help on question b
ZanzabumX [31]

The answer is A.

A positive charge’s electric field pushes out.

Hope this helps! -Avenging

8 0
3 years ago
a) A Kuiper Belt Object orbits the Sun with an average orbital distance of 40 AU. What is its orbital period? Round your answer
denpristay [2]

Answer:

253 years

Explanation:

a = Distance in Astronomical units from the sun = 40 AU

P = Orbital period in years

From Kepler's law we have

P^2=a^3

\Rightarrow P^2=40^3

\Rightarrow P^2=64000

\Rightarrow P=\sqrt{64000}

\Rightarrow P=252.982212813

\Rightarrow P\approx 253\ years

The orbital period of the Kuiper belt object is 253 years.

7 0
3 years ago
A mass of 0.40 kg is suspended on a spring which then stretches 10 cm. The mass is then removed and a second mass is placed on t
Vanyuwa [196]

Answer:

 x' = 1.01 m

Explanation:

given,

mass suspended on the spring, m = 0.40 Kg

stretches to distance, x = 10 cm  = 0. 1 m

now,

we know

m g = k x

where k is spring constant

0.4 x 9.8 = k x 0.1

  k = 39.2 N/m

now, when second mass is attached to the spring work is equal to 20 J

work done by the spring is equal to

W = \dfrac{1}{2}kx'^2

20= \dfrac{1}{2}\times 39.2\times x'^2

 x'² = 1.0204

 x' = 1.01 m

hence, the spring is stretched to 1.01 m from the second mass.

 

7 0
3 years ago
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