የ12ኛ ክፍል ውጤት ወጥቷል፤ ውጤትዎን የሚያዩበት መንገድ

የፈተና ወረቀቶች

የ12ኛ ክፍል ፊዚክስ መግቢያ ፈተና፣ 2010 ዓ.ም. (2018)

ተማሪዎች በ2010 ዓ.ም. የተፈተኑበት የ12ኛ ክፍል የፊዚክስ የዩኒቨርሲቲ መግቢያ ፈተና እውነተኛ ጥያቄዎች። እያንዳንዱ ጥያቄ ትክክለኛ መልስ እና የተብራራ ማብራሪያ አለው።

በነጻ ልምምድ ይጀምሩየጥያቄ ወረቀት

  • የተፈጥሮ ሳይንስ ዘርፍ

    ማን ይፈተናል

  • 2010 ዓ.ም. (2018)

    የፈተና ዓመት

  • 41 ጥያቄዎች ከመልስ ጋር

    በዚህ ስብስብ

ከዚህ ፈተና ጥያቄዎችን ይሞክሩ

መልሱን ከመክፈትዎ በፊት እያንዳንዱን ጥያቄ በራስዎ ይመልሱ። ማብራሪያዎቹ ወደ መማሪያ መጽሐፉ ትክክለኛ ክፍል ይጠቁማሉ፤ ስለዚህ ጥያቄ ሲስቱ የትኛውን ገጽ እንደገና ማንበብ እንዳለብዎት ያውቃሉ።

ጥያቄ 1

Four vectors are listed below A⃗=[410]\vec{A} = \begin{bmatrix} 4 \\ 10 \end{bmatrix}. B⃗=[2−5]\vec{B} = \begin{bmatrix} 2 \\ -5 \end{bmatrix}, C⃗=[−615]\vec{C} = \begin{bmatrix} -6 \\ 15 \end{bmatrix}, D⃗=[15−6]\vec{D} = \begin{bmatrix} 15 \\ -6 \end{bmatrix}, Which pair of the vectors is orthogonal?

  1. A.A⃗\vec{A} and D⃗\vec{D}
  2. B.B⃗\vec{B} and C⃗\vec{C}
  3. C.C⃗\vec{C} and D⃗\vec{D}
  4. D.A⃗\vec{A} and B⃗\vec{B}
መልሱን እና ማብራሪያውን ይመልከቱ

መልስ፦ A A⃗\vec{A} and D⃗\vec{D}

ለምን

Two vectors are orthogonal if their dot product equals zero. Calculating the dot product for A⃗\vec{A} and D⃗\vec{D} gives (4)(15)+(10)(−6)=60−60=0(4)(15) + (10)(-6) = 60 - 60 = 0, so they are orthogonal.

ጥያቄ 2

A race car is moving at a constant speed of 35m/s . A security car was moving at a speed of 5m/s as the race car passes by it and was accelerating at constant rate of 5 m/s2\mathrm{m/s^2}. What was the speed of the security car when it took over the race car ?

  1. A.35m/s
  2. B.60m/s
  3. C.5m/s
  4. D.65m/s
መልሱን እና ማብራሪያውን ይመልከቱ

መልስ፦ D 65m/s

ለምን

Equating the positions of the race car and security car: 35t=5t+12(5)t235t = 5t + \frac{1}{2}(5)t^2, which simplifies to 2.5t2=30t2.5t^2 = 30t, giving t=12 st = 12\text{ s}. The final speed of the security car is v=v0+at=5+5(12)=65 m/sv = v_0 + at = 5 + 5(12) = 65\text{ m/s}.

ጥያቄ 3

A car start moving from rest and its motion on a straight line is shown in the figure below. What is the average velocity of the car within the first 6 seconds? [Figure: Hand-drawn velocity-time graph. Vertical axis labelled v (m/s) with tick marks at 2, 4, 6, 8, 10 and 12; horizontal axis labelled t(s) with tick marks at 2, 4, 6 and 8. The trace consists of three straight segments: a line rising from the origin to a point labelled "2, 6" (t = 2 s, v = 6 m/s); a horizontal segment from that point to a point labelled "4, 6" (t = 4 s, v = 6 m/s); then a line rising from there to a point labelled "6, 10" (t = 6 s, v = 10 m/s), where the trace ends.]

  1. A.5.67m/s
  2. B.1.67 m/s
  3. C.5 m/s
  4. D.6 m/s
መልሱን እና ማብራሪያውን ይመልከቱ

መልስ፦ A 5.67m/s

ለምን

The total displacement is the area under the v-tv\text{-}t graph: Area 1 is 12(2)(6)=6 m\frac{1}{2}(2)(6) = 6\text{ m}, Area 2 is (4−2)(6)=12 m(4-2)(6) = 12\text{ m}, and Area 3 is 6+102(6−4)=16 m\frac{6+10}{2}(6-4) = 16\text{ m}, giving a total of 34 m34\text{ m}. Dividing by the total time of 6 s6\text{ s} yields an average velocity of 346≈5.67 m/s\frac{34}{6} \approx 5.67\text{ m/s}.

ጥያቄ 4

Which one of the following statements is correct regarding the motion in a plane?

  1. A.When a body moves in a horizontal circle, its velocity is constant
  2. B.When a body moves in a vertical circle, its speed is constant
  3. C.The centripetal force for a body moving in either vertical or horizontal circle is toward the center
  4. D.In projectile motion the horizontal component of the motion is uniformly accelerated motion.
መልሱን እና ማብራሪያውን ይመልከቱ

መልስ፦ C The centripetal force for a body moving in either vertical or horizontal circle is toward the center

ለምን

By definition, the centripetal force is always directed toward the center of the circular trajectory for any circular motion. Option A is incorrect because the velocity vector continuously changes direction even if speed is constant.

ጥያቄ 5

A bullet is fired at an angel of 30° above the horizontal. If air resistance is neglected, what should be the bullet's horizontal acceleration (ax⃗)(\vec{a_x}) and vertical acceleration (ay⃗)(\vec{a_y}) when it reaches its maximum height?

  1. A.Both ax⃗\vec{a_x} and ay⃗\vec{a_y} are zero
  2. B.ax⃗\vec{a_x} is 10 m/s2\mathrm{m/s^2} downward, and ay⃗\vec{a_y} is zero
  3. C.Both ax⃗\vec{a_x} and ay⃗\vec{a_y} are 10 m/s2\mathrm{m/s^2} downward
  4. D.ax⃗\vec{a_x} is zero, and ay⃗\vec{a_y} is 10 m/s2\mathrm{m/s^2} downward.
መልሱን እና ማብራሪያውን ይመልከቱ

መልስ፦ D ax⃗\vec{a_x} is zero, and ay⃗\vec{a_y} is 10 m/s2\mathrm{m/s^2} downward.

ለምን

In standard projectile motion neglecting air resistance, gravity acts solely downward with an acceleration of approximately 10 m/s210\text{ m/s}^2, while there is no force in the horizontal direction, so ax=0a_x = 0 throughout the flight.

ጥያቄ 6

Two students A and B, with masses mAm_A and mBm_B respectively, stand on a wheeled trolleys as shown in the figure below. If student A has larger mass and pushes student B, then what can be said about their accelerations? [Figure: Line drawing of two boys facing each other, each standing on a small four-wheeled trolley (skateboard-like platform) resting on a hatched horizontal ground line. The boy on the left is labelled A, the boy on the right is labelled B. A leans forward with both arms extended, his hands pressing against B's hands, so A is pushing B to the right. Both boys are drawn about the same size; no forces or numerical values are marked.]

  1. A.The acceleration of student B is less than that of student A.
  2. B.Both boys will be accelerated by the same magnitude but in opposite directions.
  3. C.The acceleration of student A is less than that of student B.
  4. D.There would be no acceleration because the forcec are equal and opposite.
መልሱን እና ማብራሪያውን ይመልከቱ

መልስ፦ C The acceleration of student A is less than that of student B.

ለምን

According to Newton's third law, the forces exerted on both students are equal in magnitude and opposite in direction. Since acceleration is inversely proportional to mass (a=F/ma = F/m), the heavier student A experiences a smaller acceleration than student B.

ሙሉውን ፈተና በ PDF ያውርዱ

በዚህ ስብስብ ያሉት 41 ጥያቄዎች ሁሉ እንደ ታትሞ የወጣ ፈተና፣ የመልስ ቁልፉም ከ39 ማብራሪያዎች ጋር እንደ ሁለተኛ ፋይል። ያትሟቸው ወይም ያለ ዳታ ያንብቧቸው።

ነጻ ነው፤ መለያም አይጠይቅም።

የጥያቄ ወረቀትየመልስ ቁልፍ እና ማብራሪያዎች

ሙሉውን ፈተና በሰዓት ተገድበው ይስሩ

ነጻ የ Temari መለያ ሙሉውን የፊዚክስ ፈተና እና የሌሎች ዓመታትንም ሁሉ በሰዓት የተገደቡ የሙከራ ፈተናዎች አድርጎ ከወዲያውኑ ውጤት ጋር ይሰጥዎታል። በማንኛውም ስልክ ላይ፣ ደካማ ግንኙነት ላይም ይሰራል።

በነጻ ልምምድ ይጀምሩ

የሌሎች ዓመታት የፊዚክስ ፈተናዎች

2018 ዓ.ም. (2026)2017 ዓ.ም. (2025)2016 ዓ.ም. (2024)2015 ዓ.ም. (2023)2014 ዓ.ም. (2022)2013 ዓ.ም. (2021)2012 ዓ.ም. (2020)2011 ዓ.ም. (2019)2009 ዓ.ም. (2017)2008 ዓ.ም. (2016)2007 ዓ.ም. (2015)2006 ዓ.ም. (2014)2005 ዓ.ም. (2013)

ይህን ፈተና ከመቀመጥዎ በፊት ፊዚክስ ያጥኑ

የ12ኛ ክፍል ፊዚክስ አጭር ማስታወሻዎች 28 ካርዶች አሏቸው። እያንዳንዱ ከአገር አቀፉ መጽሐፍ ምዕራፍ በምዕራፍ ተዘጋጅቶ ሕጉን፣ ምልክቶቹ ምን እንደሚወክሉና ውጤት የሚያሳጣውን ስህተት ይናገራል።

የፊዚክስ ማስታወሻዎችን ይክፈቱ
እነዚህ እውነተኛዎቹ የ2010 ዓ.ም. የፊዚክስ መግቢያ ፈተና ጥያቄዎች ናቸው?

አዎ። ተማሪዎች በ2010 ዓ.ም. (2018) ከተፈተኑበት ብሔራዊ የዩኒቨርሲቲ መግቢያ የፊዚክስ ፈተና የተወሰዱ ናቸው። ስብስባችን ከጥያቄዎቹ 41 ያህሉን ይዟል፤ እያንዳንዱም መልስና ወደ መማሪያ መጽሐፉ የሚጠቁም ማብራሪያ አለው።

ሙሉውን ፈተና እንዴት እለማመዳለሁ?

በስልክ ቁጥር ነጻ መለያ ይክፈቱ። ሙሉው ፈተና በሰዓት የተገደበ የሙከራ ፈተና ሆኖ ወዲያውኑ ውጤት ይሰጣል፣ በወር 199 ብር የሆነው የ AI አጋዥ ማሻሻያም ያመለጠዎትን ሁሉ ያስረዳዎታል።

የ2010 ዓ.ም. የፊዚክስ ፈተናን በ PDF ማውረድ እችላለሁ?

አዎ፤ ነጻም ነው። የጥያቄ ወረቀቱ አንድ ፋይል ነው፤ የመልስ ቁልፉ ከማብራሪያዎቹ ጋር ሌላ ፋይል ነው፤ ስለዚህ ፈተናውን መጀመሪያ መሞከር ይችላሉ። ያትሟቸው፣ ወይም ዳታ በሌለበት ጊዜ ለክለሳ በስልክዎ ላይ ያስቀምጧቸው።

ይግቡመለያዎን ይክፈቱ