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Reil [10]
3 years ago
15

The

Physics
1 answer:
yKpoI14uk [10]3 years ago
5 0

13.6g

136g

Explanation:

Given parameters:

Density of mercury = 13.6g/cm³

Unknown:

Mass of :

1cm³ of mercury

10cm³ of mercury

Solution:

Density is the mass per unit volume of a substance. It is the amount of matter in a given volume.

   Density = \frac{mass}{volume}

Since the unknown in the problem is mass;

    Mass = density x volume

Mass of  1cm³ of mercury = 13.6 x 1 = 13.6g

Mass of  10cm³ of mercury = 10 x 13.6 = 136g

learn more:

Density problems brainly.com/question/3433940

#learnwithBrainly

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Answer:

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Explanation:

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\frac{dF}{dt}=\frac{k}{r^{2}}\\\frac{dF}{dt}=k*r^{-2}\\\frac{dF}{dt}=-2*k*r^{-3} \frac{dr}{dt}\\\frac{dF}{dt}=\frac{-2k}{r^{3}}\frac{dr}{dt}

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\frac{dF}{dt}=\frac{-2*k}{r^{3}}*0.4\\\frac{dF}{dt}=\frac{-8*10x^{13}N*km^{2} }{(10x10^{4}) ^{3}} \\\frac{dF}{dt}=\frac{-8x10^{12} }{1x10^{12}}

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an athlete swims the distance from one end of a 50m pool to the other end in a time of 25 s what is The Athlete's average speed
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The basic barometer can be used to measure the height of a building. If the barometric readings at the top and at the bottom of
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h_{Hg-BOT}=695mmHg=0.695m the barometric reading at the bottom of the building

\rho _{air}=1.18 kg/m^{3} density of air

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In order to approach this problem, we will firstly use the following equations to find the pressure at the top of the building P_{TOP} and the perssure at the bottom P_{BOT}:

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P_{BOT}=\rho _{Hg} g h_{Hg-BOT} (2)

From (1): P_{TOP}=(13600 kg/m^{3})(9.8/m^{2})(0.675m)=89964 Pa (3)

From (2): P_{BOT}=(13600 kg/m^{3})(9.8/m^{2})(0.695m)=92629.6 Pa (4)

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On the other hand, we have a column of air with a cross-section area A and the same height of the building, lets name it h_{air}.

As pressure is defined as the force F exerted on a specific area A, we can write:

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\rho_{air}=\frac{m_{air}}{(A)(h_{air}} (12)

Isolating m_{air}:

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Matching (8) and (14)

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Hope that helps have a good day though.

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