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sergejj [24]
2 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]2 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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An amount of work W is done on an object of mass m initially at rest, and as result it winds up moving at speed v. Suppose inste
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Answer:

Explanation:

Given

W amount of work is done on the system such that it acquires v velocity after operation(initial velocity)

According to work energy theorem work done by all the forces is equal to change in kinetic energy of object

W=\frac{1}{2}mv^2---1

where m=mass of object

v=velocity of object

When the object is already have velocity v then the final speed is given by work energy theorem

W=\frac{1}{2}mv_f^2-\frac{1}{2}mv^2-----2

From 1 and 2 we get

\frac{1}{2}mv^2=\frac{1}{2}mv_f^2-\frac{1}{2}mv^2

2\times \frac{1}{2}mv^2=\frac{1}{2}mv_f^2

v_f^2=2v^2

v_f=\sqrt{2}v                

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2 years ago
A responder can protect himself/herself from radiation by using shielding as a response action. What materials are best for prot
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Answer:

Few millimeter thick aluminium, water, wood, acrylic glass or plastic.

Explanation:

The materials that are best for protection against beta particles are few millimeter thickness of aluminium, but for the high energy beta-particles radiations the low atomic mass materials such as plastic, wood, water and acrylic glass can be used.

These materials can also be used in personal protective equipment which includes all the clothing that can be worn to prevent any injury or illness due to the exposure to radiation.

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A particular electric car is supplied with 300 kJ of chemical energy by the battery. Of this, a total of 70.5 kJ of energy is wa
Lorico [155]

Supplied energy=300kJ

  • Wasted energy=70.5J

Used energy:-

\\ \sf\longmapsto 300-70.5=229.5kJ

We know

\boxed{\sf Efficiency=\dfrac{Used\:Energy}{Supplied\:Energy}\times 100}

\\ \sf\longmapsto Efficiency=\dfrac{229.5}{300}\times 100

\\ \sf\longmapsto Efficiency=\dfrac{229.5}{3}

\\ \sf\longmapsto Efficiency=76.5\%

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1 year ago
In a certain ideal heat engine, 10.00 kJ of heat is withdrawn from the hot source at 273 K and 3.00 kJ of work is generated. Wha
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Answer:

Temperature of the sink will be 191.1 K

Explanation:

We have given that heat withdrawn form the source = 10 KJ

Work done = 3 KJ

We know that efficiency is given by

\eta =\frac{work\ done}{heat\ withdrawn}=\frac{3}{10}=0.3

Higher temperature is given by T_1=273K

We have to find the lower temperature T_2

We know that efficiency is also given by

1-\frac{T_2}{T_1}=\frac{T_1-T_2}{T_1}

So \frac{273-T_2}{273}=0.3

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5 0
2 years ago
Read 2 more answers
A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413
natima [27]

Answer:

The value of  charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force F=14.413\ N

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

r=\sqrt{x_{2}^2+y_{2}^2}

Put the value into the formula

r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}

r=0.0876\ m

We need to calculate the magnitude of the charge q₃

Using formula of net force

F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})

Put the value into the formula

14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})

(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}

\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}

q_{3}=0.0210909\times(0.0438)^2

q_{3}=40.46\times10^{-6}\ C

q_{3}=40.46\ \mu C

Hence, The value of  charge q₃ is 40.46 μC.

5 0
2 years ago
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