የ12ኛ ክፍል ውጤት ወጥቷል፤ ውጤትዎን የሚያዩበት መንገድ

የምዕራፍ ማስታወሻዎችና የመልመጃ ጥያቄዎች

የ12ኛ ክፍል ፊዚክስ ምዕራፍ 3፦ ፍሉይድ ሜካኒክስ

በዚህ ገጽ ያለው ሁሉ ለ12ኛ ክፍል የፊዚክስ መፅሐፍ ምዕራፍ 3 ነው፦ ምዕራፉ የሚሸፍነው፣ አጭር ማስታወሻዎቹ፣ የመልመጃ ጥያቄዎች ከመልሶቻቸው ጋር፣ እና ቪዲዮ ባለበት የትምህርቱ ቪዲዮ።

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የምዕራፍ 3 የመልመጃ ጥያቄዎች
69–117
የምዕራፍ 3 የመፅሐፍ ገጾች

ምዕራፍ 3 የሚሸፍነው

  1. 1.Fluid Statics
  2. 2.Pressure in fluids at rest
  3. 3.Archimedes' principle
  4. 4.Fluid flow
  5. 5.Safety and high pressure

የምዕራፍ 3 አጭር ማስታወሻዎች

በ12ኛ ክፍል የፊዚክስ ገጽ ላይ በዚህ ምዕራፍ ሥር የተመዘገቡት ካርዶች። አንድ በአንድ ይክፈቱ።

Pabs=Patm+PgageP_{\text{abs}} = P_{\text{atm}} + P_{\text{gage}}
PabsP_{\text{abs}}
ፍጹም ግፊትPaPa
PatmP_{\text{atm}}
የአካባቢው atmospheric pressurePaPa
PgageP_{\text{gage}}
የጌጅ ግፊትPaPa

መቼ እንደሚጠቀሙበት

የመለኪያ መሳሪያ ንባብን (gauge pressure) እና ከፍፁም ባዶ ቦታ አንጻር ያለውን absolute pressure እርስ በእርስ ለመቀየር ይጠቀሙበት።

ይጠንቀቁ

የ pressure መለኪያ መሳሪያዎች ከባቢ አየር ላይ ዜሮ ያነባሉ። ideal gas equation ላይ ስሌት ሲሰሩ gauge pressure ላይ P_atm ሳይደምሩ እንዳይተኩ።

ከGrade 12 Physics፣ ገጽ 69-117 የተዘጋጀ፤ ከመውጣቱ በፊት ሁለት ጊዜ ተረጋግጧል

FB=ρfluidgVdispF_B = \rho_{\text{fluid}} g V_{\text{disp}}
FBF_B
የመንሳፈፍ ኃይልNN
ρfluid\rho_{\text{fluid}}
የፈሳሹ densitykg m−3kg\,m^{-3}
gg
በስበት ምክንያት የሚፈጠር ፍጥንጥነትm s−2m\,s^{-2}
VdispV_{\text{disp}}
የተፈናቀለው ፈሳሽ volumem3m^3

መቼ እንደሚጠቀሙበት

በፈሳሽ ውስጥ በከፊልም ሆነ ሙሉ በሙሉ በሰመጠ አካል ላይ ወደ ላይ የሚገፋውን buoyant force ለማስላት ይጠቀሙበት።

ይጠንቀቁ

የ buoyant force ስሌት ላይ የፈሳሹን density እንጂ የነገሩን density እንዳይጠቀሙ ይጠንቀቁ።

ከGrade 12 Physics፣ ገጽ 69-117 የተዘጋጀ፤ ከመውጣቱ በፊት ሁለት ጊዜ ተረጋግጧል

VdispVobj=ρobjρfluid\frac{V_{\text{disp}}}{V_{\text{obj}}} = \frac{\rho_{\text{obj}}}{\rho_{\text{fluid}}}
VdispV_{\text{disp}}
ፈሳሽ ውስጥ የሰመጠው volumem3m^3
VobjV_{\text{obj}}
የእቃው አጠቃላይ volumem3m^3
ρobj\rho_{\text{obj}}
የሚንሳፈፈው ነገር densitykg m−3kg\,m^{-3}
ρfluid\rho_{\text{fluid}}
የፈሳሹ densitykg m−3kg\,m^{-3}

መቼ እንደሚጠቀሙበት

አንድ የሚንሳፈፍ አካል ከፈሳሹ ወለል በታች ምን ያህል fraction ወይም percentage እንደሰመጠ ለማስላት ይጠቀሙበት።

ይጠንቀቁ

ይህ ቀመር የሚያገለግለው የሚንሳፈፉ አካላት ላይ ብቻ ነው፤ የነገሩ density ከፈሳሹ density ማነስ አለበት።

ከGrade 12 Physics፣ ገጽ 69-117 የተዘጋጀ፤ ከመውጣቱ በፊት ሁለት ጊዜ ተረጋግጧል

F1A1=F2A2\frac{F_1}{A_1} = \frac{F_2}{A_2}
F1F_1
በመጀመሪያው piston ላይ ያረፈ forceNN
A1A_1
የመጀመሪያው piston cross-sectional aream2m^2
F2F_2
በሁለተኛው piston ላይ የሚፈጠር forceNN
A2A_2
የሁለተኛው piston cross-sectional aream2m^2

መቼ እንደሚጠቀሙበት

የተዘጋ ፈሳሽ ጫናን በእኩል በሚያስተላልፍባቸው እንደ hydraulic lift እና hydraulic brake ባሉ መሳሪያዎች ውስጥ force ወይም area ለማስላት ይጠቀሙበት።

ይጠንቀቁ

ስሌት ከመጀመርዎ በፊት A_1 እና A_2 በተመሳሳይ unit መሆናቸውን ያረጋግጡ። diameter ከተሰጠ area ከ diameter squared ጋር ተመጣጣኝ መሆኑን ያስታውሱ።

ከGrade 12 Physics፣ ገጽ 69-117 የተዘጋጀ፤ ከመውጣቱ በፊት ሁለት ጊዜ ተረጋግጧል

P=P0+ρghP = P_0 + \rho g h
PP
depth h ላይ ያለ absolute pressurePaPa
P0P_0
የፈሳሹ surface ላይ ያለ pressurePaPa
ρ\rho
የፈሳሹ densitykg m−3kg\,m^{-3}
gg
በስበት ምክንያት የሚፈጠር ፍጥንጥነትm s−2m\,s^{-2}
hh
ከፈሳሹ ወለል በታች ያለው depthmm

መቼ እንደሚጠቀሙበት

የማይንቀሳቀስ ፈሳሽ ወለል ስር በተወሰነ ጥልቀት ላይ ያለውን ጠቅላላ hydrostatic pressure ለማስላት ይጠቀሙበት።

ይጠንቀቁ

የተጠየቀው absolute pressure ከሆነ የከባቢ አየር ጫና P_0 መጨመርን አይርሱ። gauge pressure ከተጠየቀ ግን P_0 አይካተትም።

ከGrade 12 Physics፣ ገጽ 69-117 የተዘጋጀ፤ ከመውጣቱ በፊት ሁለት ጊዜ ተረጋግጧል

የምዕራፍ 3 የመልመጃ ጥያቄዎች

በዚህ ምዕራፍ ላይ የተጻፉ 10 የምርጫ ጥያቄዎች፣ እያንዳንዳቸው መልሱና ምክንያቱ አላቸው። መልሱን ከመክፈትዎ በፊት ይሞክሩ። ጥያቄዎቹ በእንግሊዝኛ ናቸው።

ጥያቄ 1

What is the SI unit of pressure?

  1. A.Pascal (Pa)
  2. B.Atmosphere (atm)
  3. C.Millimeter of mercury (mmHg)
  4. D.Pounds per square inch (psi)
መልሱንና ምክንያቱን ያሳዩ

መልስ፦ A Pascal (Pa)

ምክንያቱ

In the SI system, pressure is measured in newtons per square meter (N/m2), which is defined as the pascal (Pa).

ጥያቄ 2

For a fluid at rest in fluid statics, what is the value of the shear (tangential) stress?

  1. A.It is equal to atmospheric pressure
  2. B.It is zero
  3. C.It increases with fluid depth
  4. D.It equals the normal stress
መልሱንና ምክንያቱን ያሳዩ

መልስ፦ B It is zero

ምክንያቱ

In a fluid at rest, there is no relative motion between adjacent fluid layers. Therefore, the shear stress is zero, and the only existing stress is the normal stress (pressure).

ጥያቄ 3

According to Pascal's principle, what happens when an external pressure is applied to an enclosed static fluid?

  1. A.The pressure decreases with distance from the source
  2. B.The pressure change is transmitted undiminished to every point in the fluid
  3. C.The pressure increases only at the bottom surface
  4. D.The fluid becomes compressed to half its volume
መልሱንና ምክንያቱን ያሳዩ

መልስ፦ B The pressure change is transmitted undiminished to every point in the fluid

ምክንያቱ

Pascal's principle states that a change in pressure applied to a static enclosed fluid is transmitted undiminished to every point of the fluid and to the walls of its container.

ጥያቄ 4

According to Archimedes' principle, the upward buoyant force acting on an object placed in a fluid equals the:

  1. A.Weight of the fluid displaced by the object
  2. B.Total weight of the object
  3. C.Surface area of the object
  4. D.Atmospheric pressure acting on the fluid surface
መልሱንና ምክንያቱን ያሳዩ

መልስ፦ A Weight of the fluid displaced by the object

ምክንያቱ

Archimedes' principle states that the buoyant force on an object equals the weight of the fluid it displaces (FB = Wfluid).

ጥያቄ 5

A mechanic in Addis Ababa measures the tire gauge pressure of a minibus to be 32.0 psi. If the local atmospheric pressure is 14.3 psi, what is the absolute pressure inside the tire?

  1. A.17.7 psi
  2. B.32.0 psi
  3. C.46.3 psi
  4. D.457.6 psi
መልሱንና ምክንያቱን ያሳዩ

መልስ፦ C 46.3 psi

ምክንያቱ

Absolute pressure is found by adding atmospheric pressure to gauge pressure: Pabs = Patm + Pgage = 14.3 psi + 32.0 psi = 46.3 psi.

ጥያቄ 6

A hydraulic press in a garage in Hawassa has a small piston with an area of 0.0020 m2 and a large piston with an area of 0.20 m2. If a total downward force of 18,000 N acts on the large piston, what minimum force must be applied to the small piston to lift the load?

  1. A.180 N
  2. B.360 N
  3. C.1,800 N
  4. D.90 N
መልሱንና ምክንያቱን ያሳዩ

መልስ፦ A 180 N

ምክንያቱ

From Pascal's principle, F1 = (A1 / A2) × F2 = (0.0020 m2 / 0.20 m2) × 18,000 N = 0.01 × 18,000 N = 180 N.

ጥያቄ 7

What is the gauge pressure in water at a depth of 25.0 m? (Take the density of water as 1.00 × 103 kg/m3 and g = 9.8 m/s2).

  1. A.2.45 &times; 10<sup>5</sup> Pa
  2. B.2.50 &times; 10<sup>4</sup> Pa
  3. C.3.46 &times; 10<sup>5</sup> Pa
  4. D.1.01 &times; 10<sup>5</sup> Pa
መልሱንና ምክንያቱን ያሳዩ

መልስ፦ A 2.45 &times; 10<sup>5</sup> Pa

ምክንያቱ

The gauge pressure due to depth is P = ρgh = (1.00 × 103 kg/m3)(9.8 m/s2)(25.0 m) = 245,000 Pa = 2.45 × 105 Pa.

ጥያቄ 8

Why does atmospheric pressure decrease as a person climbs to higher altitudes on a mountain?

  1. A.Gas particles move faster at higher altitudes
  2. B.The depth and density of the air column above both decrease
  3. C.Gravity stops acting on gas molecules at high elevation
  4. D.Water vapor disappears completely from the atmosphere
መልሱንና ምክንያቱን ያሳዩ

መልስ፦ B The depth and density of the air column above both decrease

ምክንያቱ

Atmospheric pressure is caused by the weight of the air column above. At higher altitudes, there is less depth of atmosphere above you and the air is less dense, so the atmospheric pressure decreases.

ጥያቄ 9

A rectangular wooden block floats in Lake Tana with 75% of its total volume submerged in fresh water (ρ = 1000 kg/m3). What is the density of the wooden block?

  1. A.250 kg/m<sup>3</sup>
  2. B.750 kg/m<sup>3</sup>
  3. C.1333 kg/m<sup>3</sup>
  4. D.75 kg/m<sup>3</sup>
መልሱንና ምክንያቱን ያሳዩ

መልስ፦ B 750 kg/m<sup>3</sup>

ምክንያቱ

For a floating object, Vdisp / Vobj = ρobj / ρfluid. Therefore, ρobj = 0.75 × 1000 kg/m3 = 750 kg/m3.

ጥያቄ 10

An open-end manometer is connected to a gas container. The fluid has a density of 850 kg/m3, and the liquid column in the open arm is 0.55 m higher than the arm connected to the gas. If atmospheric pressure is 96.0 kPa and g = 9.81 m/s2, what is the absolute pressure of the gas?

  1. A.91.4 kPa
  2. B.96.0 kPa
  3. C.100.6 kPa
  4. D.105.2 kPa
መልሱንና ምክንያቱን ያሳዩ

መልስ፦ C 100.6 kPa

ምክንያቱ

When the open arm is higher, Pgas = Patm + ρgh. First, ρgh = (850 kg/m3)(9.81 m/s2)(0.55 m) = 4586 Pa = 4.6 kPa. Thus, Pgas = 96.0 kPa + 4.6 kPa = 100.6 kPa.

መተግበሪያው በዚህ ምዕራፍ ላይ 12 ጥያቄዎች አሉት፤ እየሠሩ ያርምዎታል። እዚህ ያሉት የመጀመሪያዎቹ 10 ናቸው።

የምዕራፍ 3 የመፅሐፍ ገጾች

ምዕራፍ 3 በየትምህርት ሚኒስቴር እትም ከገጽ 69 እስከ ገጽ 117 ይደርሳል።

የትምህርት ሚኒስቴር እትም, እንግሊዝኛ, 186 ገጾች

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