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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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Two objects of the same size are both perfect blackbodies. One has a temperature of 3000 K, so its frequency of maximum emission
bija089 [108]

Answer:

a) The colder body (3000k), b) hearter body c) 12000K body

Explanation:

This exercise should know the power emitted by the objects and the distribution of this emission in the energy spectrum, for this we will use Stefan's laws and that of Wien's displacement

Stefan's Law                     P = σ A e T⁴

Wien displacement law   λ T = 2,898 10⁻³ m K

Let's calculate the power emitted for each object.

As they are perfect black bodies e = 1, they also indicate that they have the same area

T = 3000K

       P₁ = σ A T₁⁴

T = 12000K

       P₂ = σ A T₂⁴

       P₂ / P₁ = T₂⁴ / T₁⁴

       P₂ / P₁ = (12000/3000)⁴

       P₂ / P₁ = 256

This indicates that the hottest body emission is 256 times the coldest body emission.

Let's calculate the maximum emission wavelength

Body 1

T = 3000K

       λ T = 2,898 10-3

       λ₁ = 2.89810-3 / T

       λ₁ = 2,898 10-3 / 3000

       λ₁ = 0.966 10-6 m

      λ₁ = 966 nm

T = 12000K

      λ₂ = 2,898 10-3 / 12000

      λ₂ = 0.2415 10-6 m

      λ₂ = 214 nm

a) The colder body (3000k) emits more light in the infrared, since the emission of the hot body is at a minimum (emission tail)

b) The two bodies have emission in this region, the body of 3000K in the part of rise of the emission and the body to 12000K in the descent of the emission even when this body emits 256 times more than the other, so this body should have the highest broadcast in this area

c) The emission of the hottest 12000K body is mainly in UV

d) The hottest body emits more energy in UV and visible

e) No body has greater emission in all zones

5 0
3 years ago
A cheetah can go from the state of rest to running at 20m/s in just two seconds. What is the Cheetahs average acceleration
babymother [125]

acceleration = change in velocity/change in time

so...

a = 20 m/s / 2 seconds

a = 10


hope that helps :)

P.S. found this from Brainly User, sometimes all you have to do is search to find the answer.

7 0
3 years ago
A magnetic field is uniform over a flat, horizontal circular region with a radius of 1.50 mm, and the field varies with time. In
atroni [7]

Answer:

The average induced emf around the border of the circular region is 8.48\times 10^{-5}\ V.

Explanation:

Given that,

Radius of circular region, r = 1.5 mm

Initial magnetic field, B = 0

Final magnetic field, B' = 1.5 T

The magnetic field is pointing upward when viewed from above, perpendicular to the circular plane in a time of 125 ms. We need to find the average induced emf around the border of the circular region. It is given by the rate of change of magnetic flux as :

\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=A\dfrac{-d(B'-B)}{dt}\\\\\epsilon=\pi (1.5\times 10^{-3})^2\times \dfrac{1.5}{0.125}\\\\\epsilon=8.48\times 10^{-5}\ V

So, the average induced emf around the border of the circular region is 8.48\times 10^{-5}\ V.

6 0
3 years ago
Read 2 more answers
For a given wave if the frequency doubles the wave length?
pav-90 [236]
For a given wave in a given medium, if the frequency doubles,
the wavelength becomes 50% shorter.

That is, it becomes half as long as it was originally.
5 0
3 years ago
Which of these is not an example of
prohojiy [21]

Answer:

A) The process that give us day  and night.

Explanation:

From the answer choices provided the one that is not an example of this is the process that give us day  and night. This is because, the day/night cycle occurs as the Earth spins on it's axis. This cause one side of the Earth to be facing the Sun, which is the side that is currently experiencing day time while the other side is experiencing night time. As the Earth spins the cycles repeat. This is not an example of revolving.

5 0
3 years ago
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