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Molodets [167]
3 years ago
6

Some life forms move so slowly that their movement cannot be detected by normal observation. The best way to determine any movem

ent is by placing a mark at some point, and observing any change in position from it. Why is this true?
Physics
2 answers:
romanna [79]3 years ago
5 0
Not sure the precise concept of "normal observation", but I assume that is observed by "eyes".

Eye observation is basically macroscopic, but when you use a mark, which can be regarded as a point of mass, then it goes to microscopic.

Mark is a reference point which you can compare the relative position change, but with your eyes, first you cannot notice microscopic changes, second the eyes cannot precisely set a stable reference point.
kirza4 [7]3 years ago
3 0

The mark is a reference point.

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Hey i am here to help you.

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2 years ago
The vacuum pressure of a condenser is given to be 80 kpa. if the atmospheric pressure is 98 kpa, what is the gage pressure and a
dimaraw [331]

The absolute pressure is given by the equation,

P_{abs}=P_{atm}-P_{vac}

Here, P_{abs} is absolute pressure,P_{atm} is atmospheric pressure and P_{vac} is vacuum pressure.

Therefore,

P_{abs}=98 kPa-80 kPa=18kPa

The gage pressure is given by the equation,

P_{gage}=P_{abs}-P_{atm}.

Thus,

P_{gage}=18kPa-98 kPa=-80 kPa.

In kn/m^2,

The absolute pressure,

P_{abs}=18kPa(\frac{1kN/m^2}{kPa}) =18\ kN/m^2

The gage pressure,

P_{gage}=-80kPa(\frac{1kN/m^2}{kPa}) =-80\ kN/m^2.

In lbf/in2

The absolute pressure,

P_{abs}=18\ kPa(\frac{1.45\times 10^{-1}\ lbf/in^2 }{1kPa} )=2.6\ lbf/in^2

The gage pressure,

P_{gage}=-80kPa(\frac{1.45\times 10^{-1}\ lbf/in^2 }{1kPa} )=-11.6\ lbf/in^2

In psi,

The absolute pressure,

P_{abs}=18\ kPa(\frac{1.45037738\times 10^{-1}\ psi }{1kPa})=2.610\ psi.

The gage pressure,

P_{gage}=-80kPa(\frac{1.45037738\times 10^{-1}\ psi }{1kPa} )=-11.6030\ psi

In  mm Hg

The absolute pressure,

P_{abs}=18kPa(\frac{7.5\ mm\ of\ Hg }{1\ kPa})= 135\ mm\ of\ Hg

The gage pressure,

P_{gage}=-80kPa(\frac{7.5\ mm\ of\ Hg }{1\ kPa})=-600\ mm\ of\ Hg



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