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Fantom [35]
2 years ago
6

Plzzzzzzzzzz!!!!!!! Hurryyyyy

Physics
1 answer:
Scilla [17]2 years ago
3 0

Answer:

student  A or B

Explanation:

A common demonstration is to put a ringing alarm clock or bell in the bell jar, and when the vacuum is created, you can no longer hear the sound of the clock/bell.

The bell is connected to a lab pack or batteries and rung to show pupils it can be heard under normal circumstances. The bell jar is then connected to a vacuum pump using a vacuum plate (see Fig 2) and the air is removed from inside creating a near vacuum. The bell is then again rung. This time however, it cannot be heard.

Small low voltage buzzers can be used as a bell replacement for the bell and work in exactly the same way though teachers generally prefer bells as students may be able to see the hammer moving, proving that it is actually ringing even though they cannot hear it.

Some vacuum pumps are better than others at keeping a strong vacuum though if you cannot completely lose the sound, you will at least notice the volume decreasing.

Sound is simply a series of longitudinal waves travelling from the source, through the air to our ears. Without air present, these waves cannot form and therefore sound cannot be conveyed.

In a longitudinal wave the particles oscillate back and forth in the direction of the wave movement unlike transverse waves which like waves on the sea, single particles travel up and down and not in the direction of the wave.

Because you will not be able to create a perfect vacuum, you may still be able to hear the bell ring slightly. Vibrations from the ringing bell can also travel up to the bung in the bell jar which in turn may resonate the jar slightly. This means you may hear the bell ring, however strong the vacuum. To compensate for this, try to insulate the bell as much as possible from the bell jar. Hanging the bell using elastic cord means some of the vibrations will be absorbed by the cord and not be transferred to the bell jar.

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What force is required to accelerate a 1840 kg car from 4.77 m/s to 23.5 m/s,
neonofarm [45]
A :-) for this question , we should apply
a = v - u by t
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reviews the approach taken in problems such as this one. A bird watcher meanders through the woods, walking 0.916 km due east, 0
Verizon [17]

Answer:

Displacement: 2.230 km    Average velocity: 1.274\frac{km}{h}

Explanation:

Let's represent displacement by the letter S and the displacement in direction 49.7° as A. Displaement is a vector, so we need to decompose all the bird's displacement into their X-Y compoments. Let's go one by one:

  • 0.916 km due east is an horizontal direction and cane be seen as  direction towards the negative side of X-axis.
  • 0.928 km due south is a vertical direction and can be seen as a direction towards the negative side of Y-axis.
  • 3.52 km in a direction of 49.7° has components on X and Y  axes. It is necessary to break it down using trigonometry,

First of all. We need to sum all the X components and all the Y componets.

∑Sx = Ax -0.916 ⇒  ∑Sx = [tex]3.52cos(49.7) - 0.916

∑Sx = 1.361 km

∑Sy = Ay - 0.918 ⇒ ∑Sy = 3.52sin(49.7) - 0.918

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The total displacement is calculated using Pythagoeran therorem:

S_{total} =\sqrt{Sx^{2}+ Sy^{2} } ⇒

S_{total} = 2.230 km

With displacement calculated, we can find the average speed as follows:

V = S/t  ⇒  V = \frac{2.230}{1.750}

V = 1.274\frac{km}{h}

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