Answer:
V = 20.67 cm³
Explanation:
In this case, let's apply the Boyle's law which is:
P1V1 = P2V2
Where P1 and V1 would be condition in the water, and P2 and V2 would be the condition at the surface.
By logic, at the surface, pressure should be equals to 1 atm or 1.01x10^5 N/m²
We know the volume of the bubble at first which is 1.70 cm³ and we need to calculate V2. We know how much is P2, but we don't know the value of P1, which is the pressure of the bubble below the sea; this can be calculated using Pascal's principle which is the following expression:
P1 = Po + dgh
Where:
Po: innitial pressure, which we can assume is 1 atm
d = density of the substance, in this case, water (1000 kg/m³)
g = gravity (9.8 m/s²)
h = distance of the bubble from the surface (115 m)
Now replacing this data in the boyle's law we have the following:
P1V1 = P2V2
V2 = P1V1/P2
V2 = (Po + dgh) * V1 / P2
Replacing the data we have:
V2 = (1.01x10^5 + 1000*9.8*115) * 1.7 / 1.01x10^5
V2 = 2,087,600 / 1.01x10^5
V2 = 20.67 cm³
Elements is the simplest form of a pure substance.They cannot be broken down into any thing else ,by either physical or chemical means.In contrasts compounds are pure substances like elements ,they are made up of two or more elements.Just like elements they can be broken down into simpler substances by chemical means,but unlike elements they can only be broken down by chemical means .not both physical and chemical means like elements can.
(I hope that this was helpful.)
Answer:
98%
Explanation:
Given parameters
Mass of motor = 10kg
Height = 2m
Time = 2s
Power input = 100w
Unknown
Efficiency = ?
Solution
Efficiency is the percentage of the power output to the power input.
Power is the rate at which work is done.
Power output = mass x g x height / time
g is the acceleration due to gravity
Power output = 10x 2 x 9.8 / 2 = 98W
Efficiency = power output/ power input x 100
Efficiency = 98/100 x 100 = 98%
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.
Answer:
an energy source (AC or DC), a conductor (wire), an electrical load (device), and at least one controller (switch).
Explanation:
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