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ANEK [815]
3 years ago
8

While practicing the trumpet you notice that every time you play a particular note a window in the room rattles. How can you exp

lain this rattling in terms of wave behaviors?
Physics
1 answer:
kherson [118]3 years ago
7 0
The sound waves travel because of the pitch of the note so the window rattles.
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A long, thin superconducting wire carrying a 17 A current passes through the center of a thin, 3.0-cm-diameter ring. A uniform e
nalin [4]

Answer:

a

 \frac{dE}{dt} =-  2.72 *10^{15} \  N/C \cdot s

b

The  direction of the electric field is opposite that of the current              

Explanation:

From the question we are told that

   The current is  I  =  17\ A

   The diameter of the ring is  d =  3.0 \ cm  = 0.03 \ m

   

Generally the  radius is mathematically represented as

       r =  \frac{d}{2}

       r =  \frac{0.03}{2}

       r =  0.015 \  m

The  cross-sectional area is mathematically represented as

       A =  \pi r^2

=>     A = 3.142 *  (0.015^2)

=>    A = 7.07 *10^{-4 } \  m^ 2

Generally  according to ampere -Maxwell equation we have that

      \oint \= B \cdot  \= ds =  \mu_o  I + \epsilon_o  \mu _o\frac{ d \phi }{dt }

Now given that \= B  =  0 it implies that

     \oint \= B \cdot  \= ds = 0

So

    \mu_o  I + \epsilon_o  \mu _o\frac{ d \phi }{dt } = 0

Where  \epsilon _o is the permittivity of free space with value \epsilon_o  =  8.85*10^{-12 } \   m^{-3} \cdot kg^{-1}\cdot  s^4 \cdot A^2

            \mu_o is the permeability of free space with value  

\mu_o  =   4\pi * 10^{-7} N/A^2

      \phi is magnetic flux which is mathematically represented as

       \phi  =  E * A

Where E is the electric field strength

  So  

       \mu_o  I + \epsilon_o  \mu _o   \frac{ d [EA] }{dt }  = 0

=>   \frac{dE}{dt} =-  \frac{I}{\epsilon_o * A }

=>   \frac{dE}{dt} =-  \frac{17}{8.85*10^{-12} * 7.07*10^{-4} }

=>   \frac{dE}{dt} =-  2.72 *10^{15} \  N/C \cdot s

The  negative  sign shows that the  direction  of  the electric field is opposite that of the current

           

       

8 0
3 years ago
A cart's initial velocity is +3.0 meters per second. What is its final velocity after accelerating at a rate of 1.5 m/s2 for 8.0
Katarina [22]

Answer:

V = 15m/s

Explanation:

Given the following data;

Initial velocity = 3m/s

Time = 8secs

Acceleration = 1.5m/s²

To find the final velocity, we would use the first equation of motion;

V = U + at

Substituting into the equation, we have

V = 3 + 1.5*8

V = 3 + 12

V = 15m/s

6 0
3 years ago
3 bulbs are in series and the same 3 bulbs are in parallel with the same battery. Which bulbs will be dimmer?
salantis [7]

The bulbs connected in series will be dimmer.

<h3>What is a series connection?</h3>

When there is a connection in series, it implies that the connection is done in an end to end fashion. When the connection is in parallel the it is made in such a way that the bulbs are joined to a common junctions the same potential difference occurs across then all.

Now we know that in a series connection, the voltage across the different bulbs varies hence there is a potential drop which accounts for the decrease in the brightness of the bulb. On the other hand, the voltage across a parallel connection is the same thus the bulbs do not dim so much.

Hence, if 3 bulbs are in series and the same 3 bulbs are in parallel with the same battery, the bulbs connected in series will be dimmer.

Learn more about series connection:brainly.com/question/18713901

#SPJ1

5 0
2 years ago
The bar ab has a velocity at point a of va = [20,9] m/s and an angular velocity w = 23 rad/s. If the distance between a and b is
DiKsa [7]

speed of point A is given as

v_a = 20.9 m/s

v_a = r*\omega

here r = distance from the axis

so here we have

20.9 = r*23

r  = 0.91 m

now the distance of point A and B is 0.71 m

while the distance of axis from point A is 0.91 m

so distance of axis from point B will be given as

r = 0.91 + 0.71 = 1.62 m

now for the speed of point b is given as

v = r \omega

v = 1.62 * 23

v = 37.26 m/s

so end point B will move with speed 37.26 m/s

8 0
3 years ago
A liquid of density 1270 kg/m3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the f
FinnZ [79.3K]

Answer:

{P_2}-P_1=49.99\ KPa

Explanation:

v_1=9.43\ m/s

d_1=11.7\ cm/s

d_2=17.5\ cm/s

From continuity equation

A_1v_1=A_2v_2

v_1d_1^2=v_2d_2^2

v_2=\dfrac{9.43\times 11.7^2}{17.5^2}

v_2=4.21\ m/s

\dfrac{P_1}{\rho g}+\dfrac{v_1^2}{2g}+y_1=\dfrac{P_2}{\rho g}+\dfrac{v_2^2}{2g}+y_2

{P_1}+\rho\dfrac{v_1^2}{2}+\rho y_1g={P_2}+\rho\dfrac{v_2^2}{2}+\rho y_2g

\rho\dfrac{v_1^2}{2}+\rho y_1g -\rho\dfrac{v_2^2}{2}-\rho y_2g={P_2}-P_1

1270\times \dfrac{9.43^2}{2}+1270\times 0\times 10 -1270\times\dfrac{4.21^2}{2}-1270\times 0.175\times 10={P_2}-P_1

{P_2}-P_1=49.99\ KPa

4 0
3 years ago
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