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.
Explanation:
If the size and direction of the forces on the object are exactly balanced , then there is no net force acting on the object
Answer:
Capacitance of the second capacitor = 2C
Explanation:
Where A is the area, d is the gap between plates and ε₀ is the dielectric constant.
Let C₁ be the capacitance of first capacitor with area A₁ and gap between plates d₁.
We have
Similarly for capacitor 2
Capacitance of the second capacitor = 2C
Answer:
Pressure, P = 32666.66 Pa
Explanation:
It is given that,
Surface area of foot of Bimaba is 150 cm² or 0.015 m².
Her weight is 50 kg
We need to find the pressure does she exert on the ground, as she stands on her one foot. The force acting per unit area is called pressure. It can be given by :
So, the pressure is 32666.66 Pa.
