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oksano4ka [1.4K]
3 years ago
5

Bernoulli’s principle helps explain the hydraulics of brake fluid in cars. true or false

Physics
2 answers:
IceJOKER [234]3 years ago
8 0
This statement is true.

<span>Bernoulli’s principle explains the relation between the speed of the fluid and the pressure or potential energy of the fluid.
Bernoulli explained that </span>an increase in the speed of a fluid is accompanied with a decrease in pressure<span> and potential energy of the fluid.  

This principle can be applied to the hydraulics </span><span>of brake fluid in cars. </span>
gavmur [86]3 years ago
7 0

Explanation :

The principle behind Hydraulic brakes is Pascal's law.

According to Bernoulli's principle, the speed of a fluid depends inversely on the pressure or on the potential energy of the fluid.

Mathematically,

P_1+\dfrac{1}{2}\rho v_1^2+\rho gh_1=P_2+\dfrac{1}{2}\rho v_2^2+\rho gh_2

According to Pascal's law, there is an equal distribution of the pressure in the container having fluids i.e.

P_1=P_2

Hence, the given statement is false.

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8 miles = 12.87 km
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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
Elan Coil [88]

Answer: 230.50 m

Explanation:

We have the following information:

h_{Hg-TOP}=675mmHg=0.675m the barometric reading at the top of the building

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

\rho _{Hg}=13600 kg/m^{3} density of mercury

g=9.8/m^{2} gravity

And we need to find the height of the building.

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}:

P_{TOP}=\rho _{Hg} g h_{Hg-TOP} (1)

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)

Having the pressures at the top and the bottom of the building, we can calculate the variation in pressure \Delta P:

\Delta P=P_{BOT} - P_{TOP} (5)

\Delta P=92629.6 Pa - 89964 Pa=2665.6 Pa (6)

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:

\Delta P=\frac{F}{A} (7)

If we isolate F we have:

F= A \Delta P (8)

Also, the force gravity exerts on this column of air (its weight) is:

F=m_{air} g (9)

Knowing the density of air is: \rho_{air}=\frac{m_{air}}{V_{air}} (10)

where the volume of air can be written as: V_{air}=(A)(h_{air}) (11)

Substituting (1) in (10):

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

Isolating m_{air}:

m_{air}=(\rho_{air}) (A) (h_{air}) (13)

Substituting (13) in (9):

F=(\rho_{air}) (A) (h_{air}) (g) (14)

Matching (8) and (14)

A \Delta P=(\rho_{air}) (A) (h_{air}) (g) (15)

Isolating h_{air}:

h_{air}=\frac{\Delta P}{g \rho_{air}} (16)

Substituting the known and calculated values:

h_{air}=\frac{2665.6 Pa}{(9.8m/s^{2}) (1.18 kg/m^{3})} (17)

Finally:

h_{air}=230.50 m This is the height of the building

8 0
3 years ago
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