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lukranit [14]
3 years ago
15

In an experiment, a ringing bell is placed in a vacuum jar that does not have any air in it. What best describes why the bell is

seen vibrating but not heard? Light waves have high frequencies and are able to pass through the vacuum jar, but the frequencies of sound waves are too low to vibrate a vacuum. Light waves travel faster than sound waves, so they are able to travel through the vacuum jar, but sound waves do not have enough energy to vibrate glass. Light does not need a medium to travel through, but since sound waves must have a medium to vibrate, sound is not created where no air is present. Light is a longitudinal wave that is able to pass through the vacuum jar, but sound is a transverse wave that has to move perpendicular to air.
Physics
2 answers:
Neporo4naja [7]3 years ago
4 0
I believe the correct answer from the choices listed above is the third option. The bell is seen vibrating but not heard because l<span>ight does not need a medium to travel through, but since sound waves must have a medium to vibrate, sound is not created where no air is present. Hope this answers the question. Have a nice day.</span>
Eva8 [605]3 years ago
4 0

Light does not need a medium to travel through, but since sound waves must have a medium to vibrate, sound is not created where no air is present

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A positive charge of 18nC is evenly distributed along a straight rod of length 4.0 m that is bent into a circular arc with a rad
sergey [27]

Explanation:

Formula for angle subtended at the center of the circular arc is as follows.

           \theta = \frac{S}{r}

where,   S = length of the rod

              r = radius

Putting the given values into the above formula as follows.      

                \theta = \frac{S}{r}

                             = \frac{4}{2}

                             = 2 radians (\frac{180^{o}}{\pi})

                             = 114.64^{o}

Now, we will calculate the charge density as follows.

                 \lambda = \frac{Q}{L}

                            = \frac{18 \times 10^{-9} C}{4 m}

                            = 4.5 \times 10^{-9} C/m

Now, at the center of arc we will calculate the electric field as follows.

                 E = \frac{2k \lambda Sin (\frac{\theta}{2})}{r}

                     = \frac{2(9 \times 10^{9} Nm^{2}/C^{2})(4.5 \times 10^{-9}) Sin (\frac{114.64^{o}}{2})}{2 m}

                      = 34.08 N/C

Thus, we can conclude that the magnitude of the electric eld at the center of curvature of the arc is 34.08 N/C.

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3 years ago
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Let's assume that ground level is the height 0 meters. The change in potential energy is going to be gravitational potential energy, which is given by PE=mgh.
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saw5 [17]
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3 0
2 years ago
You are looking down on a N = 9 turn coil in a magnetic field B = 0.5 T which points directly down into the screen. If the diame
Novay_Z [31]

Answer:

The magnitude of the voltage is 2.27\times10^{-4}\ V and the direction of the current is clockwise.

Explanation:

Given that,

Number of turns = 9

Magnetic field = 0.5 T

Diameter = 3 cm

Time t = 0.14 s

We need to calculate the flux

Using formula of flux

\phi=NAB

Put the value into the formula

\phi=9\times\pi\times(1.5\times10^{-2})^2\times0.5

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We need to calculate the emf

Using formula of emf

\epsilon=-\dfrac{d\phi}{dt}

\epsilon=-\dfrac{0.003180}{0.14}

\epsilon =-0.000227\ V

\epsilon=-2.27\times10^{-4}\ V

Negative sign shows the direction of current.

Hence, The magnitude of the voltage is 2.27\times10^{-4}\ V and the direction of the current is clockwise.

8 0
3 years ago
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