Physics;Flow rate, high school

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Problem# h2phy1 - Pressure converted inside the Aorta

Problem# h2phy2 - Pressure inside vein

Problem# h2phy3 - Pressure on the floor

Problem# h2phy4 - Pressure under the sea

Problem# h2phy5 - Pressure between oil and water

Problem# h2phy6 - Buoyant Force

Problem# h2phy7 - Flow rate; area vs velocity

The stream of water emerging from a faucet
"necks down" as it falls. The cross sectional area
A_o  is 1.2 cm², and A is 0.35 cm².  The two levels
are seperated by a vertical distance h=45 mm.
At what rate does water flow from the tap?

Strategy - the rule "Must have" of Mr.Zhang ®

Flow Rate Equations;
Flow Rate (Q)	=	Area x Velocity     → (1)
	=	AV
V2	=	U2+2gs     → (2)
	=	Vo2+2gs

Solution
Compute the initial velocity ;
From equation (1) ;
AoVo	=	AV

AoVo A

=V     → (3)

We bring equation(3) substitue in equation(2) ;

The Volume flow rate is [R]	=	AoVo
R	=	1.2 cm2x28.6 cm/s
Ans.  Water flow from the tap	=	34 cm3/s

Water falls from a tap,whether its speed increases or not?

If "yes", why?

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physics fluid buoyant force

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Problem# h2phy1 - Pressure converted inside the Aorta

Problem# h2phy2 - Pressure inside vein

Problem# h2phy3 - Pressure on the floor

Problem# h2phy4 - Pressure under the sea

Problem# h2phy5 - Pressure between oil and water

Problem# h2phy6 - Buoyant Force

The density of fluid "A" is 1.2 times of density fluid "B".
When we bring an object lower into the liquid "B", and
find that volume of the object sunk down is 0.6 times of total volume.
If we take this object lower into the liquid "A", and sunk down, what is
the ratio of lower portion into liquid "A" comparing with the total volume?

Strategy - the rule "Must have" of Mr.Zhang ®

Fb= ρvg

Solution

Consider, force acts against object when 1st. lowering into the “B” liquid ;

ΣFup=ΣFdown

Fb=mg

ρbg(0.6V)=mg

ρb(0.6V)=m     → (1)

Consider, force acts against object when 2nd. lowering into the “A” liquid ;

ΣFup=ΣFdown

Fb=mg

We represents Vx  with the portion which sunk down into liquid "A"

ρag(vx)=mg     → (2)

From the problem gives ρa=1.2ρb,and we also take the equation(1) substitue in equation(2) , therefore,

1.2ρbg(vx)=0.6ρbvg

1.2ρb(vx)=0.6ρbv

1.2(vx)=0.6v

vx= 0.6v 1.2

vx=0.5v

Ans.  The ratio of lower portion into liquid "A" comparing with the total volume is 0.5v

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Pressure between oil and water

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Problem# h2phy1 - Pressure converted inside the Aorta

Problem# h2phy2 - Pressure inside vein

Problem# h2phy3 - Pressure on the floor

Problem# h2phy4 - Pressure under the sea

Problem# h2phy5 - Pressure between oil and water

The U-tube contains water of density&nbspρ_w in the right arm
and oil of unknown density ρ_x in the left.   Measurement gives
L = 135 mm. and d = 12.3 mm. What is the density of the oil ?

Strategy - the rule "Must have" of Mr.Zhang ®

Interface area is joint of the two substances. P = (Po) + (Pg) Absolute pressure = Atmospheric pressure + Gauge pressure       Material or Object         Density (kg/m3)     Water: 20 οC and 1 atm 0.998x103   1 atm = 760 mmHg = 760 torr = 29.9 in.Hg =101.325 kPa = 14.7 lb/in.2

Solution Pint = Po +&nbspρwgl          (right side)  →(1) Pint = Po +&nbspρoilg(d+l)    (left side)  →(2) (1)=(2);   Po +&nbspρwgl = Po +&nbspρoilg(d+l)     ρwgl = ρoilg(d+l)     ρwl = ρoil(d+l)     (ρw)/(d+l) = ρoil ∴The density of the oil is 916.5 kg/m3

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Pressure under the water

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Problem# h2phy1 - Pressure converted inside the Aorta

Problem# h2phy2 - Pressure inside vein

Problem# h2phy3 - Pressure on the floor

Problem# h2phy4 - Pressure under the sea

An enterprising diver reasons that if a typical 20 cm.
long snorkel tube works, a 6.0 m. long tube should
also. If he foolishly uses such a tube, what is the pressure
different Δp between the external pressure
on him and the air pressure in his lungs? Why is he in danger?

Strategy - the rule "Must have" of Mr.Zhang ®

Absolute pressure (P) The difference between absolute pressure and atmospheric pressure is called gauge pressure. P =  (Pg) + (Po) Absolute pressure - Atmospheric pressure = Gauge pressure       Material or Object         Density (kg/m3)     Water: 20 οC and 1 atm 0.998x103 Sea water: 20 οC and 1 atm 1.024x103   1 atm = 760 mmHg = 760 torr = 29.9 in.Hg =101.325 kPa = 14.7 lb/in.2

Solution The diver at depth L = 6.0 m without the snorkel tube. The external pressure on him is given by ; P =  Po + ρgh ΔP=P-Po=ρgL ; because L=h     =(1000 kg/m3)(9.8 m/s2)(6.0m)     =5.9x104 Pa ∴The pressure difference is about 0.6 atm, is sufficient to collapse the lungs and force the still pressurized blood into them,a process known lung squeeze. "thoracic squeeze, also called Lung Squeeze, compression of the lungs and thoracic (chest) cavity that occurs during a breath-holding dive under water. During the descent, an increase in pressure causes air spaces and gas pockets within the body to compress." credited : Britannica

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Physics-High-school-Fluid

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Problem# h2phy1 - Pressure inside the Aorta

Problem# h2phy2 - Pressure on the floor

A living room has floor dimension of 3.5 m and 4.2 m and height of 2.4 m.
a. What does the air in the room weight?
b. What force does the atmosphere exert on the floor of the room?

Strategy - the rule "Must have" of Mr.Zhang ® material for solving this problem ; (1) w=mg    (2) m=ρv    (3) ρ=F⊥/A

Material or Object         Density (kg/m3)     Air: 20 οC and 1 atm 1.21

Solution a) The air in the room weighs;  w=mg    →(1)  m=ρv    →(2) We take (2) substitue in (1) ∴  W=ρairvg ∴  W = (1.21 kg/m3)(3.5 m x 4.2 m x 2.4 m)(9.8 m/s2 ∴  W = 418 kg m/s2 ∴  W = 418 N

Ans.   ∴The air in the room weighs 418 N

b) What force does the atmosphere exert on the floor of the room?  ρ=F⊥/A  ρatm=F⊥/A  F⊥=ρatmA  F⊥=1.01x105 N/m2)(3.5 m x4.2 m)  F⊥=1.5x106 N

Ans.   ∴The atmosphere exerts 1.5x106 N on the floor of the floor Brake Booster

Physics high school-fluid

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Problem# h2phy2 - Pressure on the floor
Problem# h2phy1 - Blood Pressure inside the Aorta

The average gauge pressure in the human aorta is about 100 mmHg.
Convert this average blood pressure to pascals.

Strategy - the rule "Must have" of Mr.Zhang ® memorized (1)Atmospheric pressure,(2)Gauge pressure,(3)Absolute pressure.

Atmospheric pressure (Pa)

Barometer is a measurement tool for measuring air pressure. Barometers measure the current air pressure at a particular location in "inches of mercury" or in "millibars" (mb).

How much pressure are you under? Earth's atmosphere is pressing against each square inch of you with a force of 1 kilogram per square centimeter (14.7 pounds per square inch).

1 atm is calculated by formulae pHggh    →  pHg=13.6x10 3 kg/m3 Therefore, 13.6x103 kg/m3x9.8 m/s2x0.76 m.=1.01325x105Nm-2 =1.01325x105Pa = 1 atm

1 atm = 760 mmHg = 760 torr = 29.9 in.Hg =101.325 kPa = 14.7 lb/in.2

Gauge pressure (Pg) is the pressure according to the liquid substance only,at the height level point "h", and it can be calculated by the liquid weight. We calculate the gauge pressure with pgh

Absolute pressure (P) The difference between absolute pressure and atmospheric pressure is called gauge pressure.

P =  (Pg) + (Pa)

Absolute pressure - Atmospheric pressure = Gauge pressure

Solution

Ans.   The average blood pressure is converted to 13.3 kPa

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brake-shoe-physics-ฟิสิกส์มัธยมปลาย-ฟิสิกส์ ม.5

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Problem h2b1: Brake system;Disc brake

Problem h2b2: Brake system;Drum brake

Consider the automobile hydraulic system shown in Figure a
A force of 100 N is applied to the brake pedal, which acts on the cylinder-called
the master-through a lever. The master cylinder has a diameter
0.500 cm. The slave cylinder (Drum brake) has a diameter 2.50 cm.
Calculate the force at each of the slave cylinders.

Strategy

Force is increased by the simple lever, and again by the hydraulic system.

Pascal's law - pressure applied anywhere to a body of fluid causes a force to be transmitted equally in all directions; the force acts at right angles to any surface in contact with the fluid; "the hydraulic press is an application of Pascal's law"

Solution

1. We take "O" to be fulcrum to find F₁

Then 500 N. is exerted on the master cylinder.
Pressure created in the master cylinder is transmitted to four so-called slave cylinders.
Then we calculate the Force at the slave cylinders each.
The circle cross sectional area of master and slave cylinder are A₁  A₂  respectively.
Hence we can find the force  F₂  at the brake drum.

Ans.   The force at each of the slave cylinders is 1.25 x 10⁴ N.

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