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sergejj [24]
3 years ago
13

A small sphere of radius R is arranged to pulsate so that its radius varies in simple harmonic motion between a minimum of R−x a

nd a maximum of R+x with frequency f. This produces sound waves in the surrounding air of density rho and bulk modulus B.
a- Find the intensity of sound waves at the surface of the sphere. (The amplitude of oscillation of the sphere is the same as that of the air at the surface of the sphere.)


b-Find the total acoustic power radiated by the sphere


c-At a distance d≫R from the center of the sphere, find the amplitude of the sound wave.


d-At a distance d≫R from the center of the sphere, find the pressure amplitude of the sound wave.


e-At a distance d≫R from the center of the sphere, find the intensity of the sound wave.


Express your answer in terms of the variables R, x, f, and appropriate constants.
Physics
1 answer:
Colt1911 [192]3 years ago
5 0

Answer:

The intensity of sound wave at the surface of the sphere I =   \frac{ 2\pi^{2}R^{2} f^{2}\sqrt{\rho B}(\triangle R)^{2}}{ d^{2} }

Explanation:

B = Bulk modulus

Intensity, I = \frac{P_{max} ^{2} }{2\sqrt{\rho B} }

The amplitude of oscillation of the sphere is given by:

P_{max} = BkA\\k = \frac{2\pi }{\lambda} \\

A = \triangle R\\

Substitute v and A into Pmax

P_{max} = (2\pi f)\sqrt{\rho B} \triangle R\\ P_{max} ^{2} = 4\pi^{2} *f^{2} \rho B (\triangle R)^{2}

I = \frac{ 4\pi^{2} f^{2} \rho B (\triangle R)^{2}}{2\sqrt{\rho B} }

P_{total} = 4\pi R^{2} I

P_{total} =4\pi R^{2}  \frac{ 2\pi^{2} f^{2} \rho B (\triangle R)^{2}}{\sqrt{\rho B} }

The intensity of the sound wave at a distance  is given by:

I = \frac{P_{total} }{4\pi d^{2} }

I = 4\pi R^{2}  \frac{ 2\pi^{2} f^{2} \rho B (\triangle R)^{2}}{\sqrt{\rho B} } * \frac{1}{4\pi d^{2} } \\I =   \frac{ 2\pi^{2}R^{2} f^{2}\sqrt{\rho B}(\triangle R)^{2}}{ d^{2} }

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At which of the following points does a roller coaster have the most potential energy? As it is going down a hill. At the top of
vova2212 [387]

Answer:

At the top of the hill.

Explanation:

As the roller coaster goes up the hill, kinetic energy (K.E) decreases, gravitational potential energy (G.P.E) increases .

As it reach the top of the hill, K.E becomes zero and G.P.E reaches <em>m</em><em>a</em><em>x</em><em>i</em><em>m</em><em>u</em><em>m</em> .

As it goes down the hill, K.E starts to increase and G.P.E decrease .

At the bottom of the hill, K.E reaches <em>maximum</em> and G.P.E becomes zero .

(Correct me it I am wrong)

6 0
3 years ago
A cyclist is riding a bicycle at a speed of 22 mph on a horizontal road. The distance between the axles is 42 in., and the mass
stealth61 [152]

Answer:

The shortest distance is  S = 24.86 ft

Explanation:

The free body diagram of this question is shown on the first uploaded image

From the question we are told that

   The speed of the bicycle is v_b = 22\ mph = 22 * \frac{5280}{3600}   =  32.26 ft/s

     The distance between the axial is  d = 42 \ in

The mass center of the cyclist and the bicycle is m_c = 26 \ in  behind the front axle

The mass center of the cyclist and the bicycle is m_h = 40 \ in above the ground

   For the bicycle not to be thrown over the

     Momentum about the back wheel must be zero so

                \sum _B = 0

=>             mg (26) = ma(40)

=>             a = \frac{26}{40} g

Here  g = 32.2 ft/s^2

     So     a =  \frac{26}{40} (32.2)

             a =  20.93 ft/s^2

Apply the equation of motion to this motion we have

       v^2 = u^2 + 2as

 Where  u = 32.26 ft /s

             and v = 0 since the bicycle is coming to a stop

        v^2 = (32.26)^2 - 2(20.93) S

=>      S = 24.86 ft        

                 

5 0
2 years ago
A cyclist is riding his bike up a mountain trail. When he starts up the trail, he is going 8 m/s. As the trail gets steeper, he
shusha [124]

Answer:

a)   a = - 0.0833 m / s²,  b)   t = 4.4 s

Explanation:

a) this is a kinematics exercise where the acceleration is along the inclined plane

         v = v₀ - a t

         a = v₀ - v / t

         a = 3 - 8/60

         a = - 0.0833 m / s²

b) in this case the final velocity is zero

         v = v₀ - a t

         0 = v₀ - at

         t = v₀ / a

         t = 28 / 6.4

         t = 4.375 s

         t = 4.4 s

3 0
2 years ago
Bill throws a tennis ball to his dog. He throws the ball at a speed of 15 m/s at an angle of 30° to the horizontal. Assume he th
Sidana [21]

1a) Bill and the dog must have a speed of 13.0 m/s

1b) The speed of the dog must be 22.5 m/s

2a) The ball passes over the outfielder's head at 3.33 s

2b) The ball passes 1.2 m above the glove

2c) The player can jump after 2.10 s or 3.13 s after the ball has been hit

2d) One solution is when the player is jumping up, the other solution is when the player is falling down

Explanation:

1a)

The motion of the ball in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

In part a), we want to know at what speed Bill and the dog have to run in order to intercept the ball as it lands on the ground: this means that Bill and the dog must have the same velocity as the horizontal velocity of the ball.

The ball's initial speed is

u = 15 m/s

And the angle of projection is

\theta=30^{\circ}

So, the ball's horizontal velocity is

v_x = u cos \theta = (15)(cos 30)=13.0 m/s

And therefore, Bill and the dog must have this speed.

1b)

For this part, we have to consider the vertical motion of the ball first.

The vertical position of the ball at time t is given by

y=u_yt+\frac{1}{2}at^2

where

u_y = u sin \theta = (15)(sin 30) = 7.5 m/s is the initial vertical velocity

a=g=-9.8 m/s^2 is the acceleration of gravity

The ball is at a position of y = 2 m above the ground when:

2=7.5t + \frac{1}{2}(-9.8)t^2\\4.9t^2-7.5t+2=0

Which has two solutions: t=0.34 s and t=1.19 s. We are told that the ball is falling to the ground, so we have to consider the second solution, t = 1.19 s.

The horizontal distance covered by the ball during this time is

d=v_x t =(13.0)(1.19)=15.5 m

The dog must be there 0.5 s before, so at a time

t' = t - 0.5 = 0.69 s

So, the speed of the dog must be

v_x' = \frac{d}{t'}=\frac{15.5}{0.69}=22.5 m/s

2a)

Here we just need to consider the horizontal motion of the ball.

The horizontal distance covered is

d=98 m

while the horizontal velocity of the ball is

v_x = u cos \theta = (34)(cos 30)=29.4 m/s

where u = 34 m/s is the initial speed.

So, the time taken for the ball to cover this distance is

t=\frac{d}{v_x}=\frac{98}{29.4}=3.33 s

2b)

Here we need to calculate the vertical position of the ball at t = 3.33 s.

The vertical position is given by

y= h + u_y t + \frac{1}{2}at^2

where

h = 1.2 m is the initial height

u_y = u sin \theta = (34)(sin 30)=17.0 m/s is the initial vertical velocity

a=g=-9.8 m/s^2 is the acceleration of gravity

Substituting t = 3.33 s,

y=1.2+(17)(3.33)+\frac{1}{2}(-9.8)(3.33)^2=3.5 m

And sinc the glove is at a height of y' = 2.3 m, the difference in height is

y - y' = 3.5 - 2.3 = 1.2 m

2c)

In order to intercept the ball, he jumps upward at a vertical speed of

u_y' = 7 m/s

So its position of the glove at time t' is

y'= h' + u_y' t' + \frac{1}{2}at'^2

where h' = 2.3 m is the initial height of the glove, and t' is the time from the moment when he jumps. To catch the ball, the height must be

y' = y = 3.5 m (the height of the ball)

Substituting and solving for t', we find

3.5 = 2.3 + 7t' -4.9t'^2\\4.9t'^2-7t'+12 = 0

Which has two solutions: t' = 0.20 s, t' = 1.23 s. But this is the time t' that the player takes to reach the same height of the ball: so the corresponding time after the ball has been hit is

t'' = t -t'

So we have two solutions:

t'' = 3.33 s - 0.20 s = 3.13 s\\t'' = 3.33 s - 1.23 s = 2.10 s

So, the player can jump after 2.10 s or after 3.13 s.

2d)

The reason for the two solutions is the following: the motion of the player is a free fall motion, so initially he jump upwards, then because of gravity he is accelerated downward, and therefore eventually he reaches a maximum height and then he  falls down.

Therefore, the two solutions corresponds to the two different part of the motion.

The first solution, t'' = 2.10 s, is the time at which the player catches the ball while he is in motion upward.

On the other hand, the second solution t'' = 3.13 s, is the time at which the player catches the ball while falling down.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

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3 years ago
Something that has many particles in a small space would have a ————- density
Diano4ka-milaya [45]

Answer:

a high density i believe

Explanation:

4 0
2 years ago
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