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Korvikt [17]
3 years ago
15

The intensity at distance from a spherically symmetric sound source is 100 W/m2. What is the intensity at five times this distan

ce from the source?
Physics
1 answer:
ss7ja [257]3 years ago
3 0

To solve this problem it is necessary to apply the concepts related to intensity as a function of power and area.

Intensity is defined to be the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity I is

I = \frac{P}{A}

The area of a sphere is given by

A = 4\pi r^2

So replacing we have to

I = \frac{P}{4\pi r^2}

Since the question tells us to find the proportion when

r_1 = 5r_2 \rightarrow \frac{r_2}{r_1} = \frac{1}{5}

So considering the two intensities we have to

I_1 = \frac{P_1}{4\pi r_1^2}

I_2 = \frac{P_2}{4\pi r_2^2}

The ratio between the two intensities would be

\frac{I_1}{I_2} = \frac{ \frac{P_1}{4\pi r_1^2}}{\frac{P_2}{4\pi r_2^2}}

The power does not change therefore it remains constant, which allows summarizing the expression to

\frac{I_1}{I_2}=(\frac{r_2}{r_1})^2

Re-arrange to find I_2

I_2 = I_1 (\frac{r_1}{r_2})^2

I_2 = 100*(\frac{1}{5})^2

I_2 = 4W/m^2

Therefore the intensity at five times this distance from the source is 4W/m^2

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(#1). (D).

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A solid sphere of radius R is placed at a height of 30 cm on a15 degree slope. It is released and rolls, without slipping, to th
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Answer:

The height is  h_c = 42.857

A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter )

Explanation:

   From the question we are told that

           The height is h_s = 30 \ cm

            The angle of the slope is \theta = 15^o

According to the law of conservation of energy

     The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

                          mgh = \frac{1}{2} I w^2 + \frac{1}{2}mv^2

Where I is the moment of inertia which is mathematically represented as this for  a sphere

                    I = \frac{2}{5} mr^2

  The angular velocity w is mathematically represented as

                         w = \frac{v}{r}

So the equation for conservation of energy becomes

               mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2

              mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]

             mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]

            gh_s =[\frac{7}{10} ] v^2

              v^2 = \frac{10gh_s}{7}

Considering a circular hoop

   The moment of inertial is different for circle and it is mathematically represented as

             I = mr^2

Substituting this into the conservation equation above

              mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2

Where h_c is the height where the circular hoop would be released to equal the speed of the sphere at the bottom

                 mgh_c  = mv^2

                     gh_c = v^2

                     h_c = \frac{v^2}{g}

Recall that   v^2 = \frac{10gh_s}{7}

                    h_c= \frac{\frac{10gh_s}{7} }{g}

                      = \frac{10h_s}{7}

            Substituting values

                   h_c = \frac{10(30)}{7}

                       h_c = 42.86 \ cm    

       

     

                         

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