Yaadannoo boqonnaa fi gaaffii shaakalaa

Boqonnaa 2 Fiizikii kutaa 12

Wanti fuula kana irra jiru hundi boqonnaa 2 kitaaba Fiizikii kutaa 12 tiif: wanta boqonnaan haguugu, yaadannoo gabaabaa, gaaffii shaakalaa deebii isaanii waliin, fi viidiyoo barnootaa bakka jirutti.

5
Kaardii: 10
Gaaffii shaakalaa boqonnaa 2: 25–68
Fuula kitaabaa boqonnaa 2

Boqonnaan 2 maal haguuga

1.Projectile motion
2.Rotational Motion
3.Rotational Dynamics
4.Planetary motion and Kepler's laws
5.Newton's law of universal Gravitation

Yaadannoo gabaabaa boqonnaa 2

Kaardiiwwan fuula Fiizikii kutaa 12 irratti boqonnaa kana jala galmaa'an. Tokko tokkoon bani.

Kepler Third Law and Orbital PeriodFoormulaa

\frac{T^2}{r^3} = \frac{4\pi^2}{G M}

\frac{T^2}{r^3} = \frac{4\pi^2}{G M}

$T$: Yeroo orbiitii $s$
$r$: Raadiyeesii orbiitii $m$
$G$: Dhaabbataa harkisa yuunivarsaalii $N\,m^2\,kg^{-2}$
$M$: Hanga qaama giddu galeessaa $kg$

Yoom akka itti fayyadamtu

Yeroo naanna'uu orbital period fi orbital radius pilaaneetii ykn saatalaayitii walitti hidhuuf fayyadamaa.

Of eeggadhu

Ulfaatinni M ulfaatina qaama giddugaleessaa kan naannatamuuti malee kan saatalaayitii miti.

Grade 12 Physics, fuula 25-68 irraa qophaa'e; osoo hin ba'in dura yeroo lama mirkanaa'e

Newton Law of Universal GravitationFoormulaa

F_g = \frac{G m_1 m_2}{r^2}

F_g = \frac{G m_1 m_2}{r^2}

$F_g$: Humna harkisaa $N$
$G$: Dhaabbataa harkisa yuunivarsaalii $N\,m^2\,kg^{-2}$
$m_1$: Hanga qaama jalqabaa $kg$
$m_2$: Hanga qaama lammaffaa $kg$
$r$: Fageenya gidduu wiirtuuwwan hangaa $m$

Yoom akka itti fayyadamtu

Humni harkisa lafaa qaamolee ulfaatina qaban lamaan gidduu jiru shallaguuf fayyadamaa.

Of eeggadhu

Fageenyi r wiirtuu qaamolee lamaanii gidduutti safarama malee fuula isaanii gidduutti miti.

Grade 12 Physics, fuula 25-68 irraa qophaa'e; osoo hin ba'in dura yeroo lama mirkanaa'e

Projectile Motion RangeFoormulaa

R = \frac{v_0^2 \sin 2\theta}{g}

R = \frac{v_0^2 \sin 2\theta}{g}

$R$: Fageenya dalgee $m$
$v_0$: Ariitii jalqabaa $m\,s^{-1}$
$\theta$: Kofa darbinsaa $rad$
$g$: Saffisiisa harkisa lafaa $m\,s^{-2}$

Yoom akka itti fayyadamtu

Yeroo projectile lafa olka'iinsa walqixaa irraa ka'ee fi irratti qubatuuf horizontal range shallaguuf fayyadamaa.

Of eeggadhu

Foormulaan kun kan hojjatu yoo olka'iinsi ka'umsaa fi qubannaa walqixa ta'e qofaadha.

Grade 12 Physics, fuula 25-68 irraa qophaa'e; osoo hin ba'in dura yeroo lama mirkanaa'e

Rotational Kinematics AnalogyGabatee wabiiv_f = v_0 + a t · Δs = v_0 t + 1/2 a t^2 · v_f^2 = v_0^2 + 2 a Δs · +1

Linear motion	Rotational motion
$v_f = v_0 + a t$	$\omega_f = \omega_0 + \alpha t$
$\Delta s = v_0 t + \frac{1}{2} a t^2$	$\Delta\theta = \omega_0 t + \frac{1}{2}\alpha t^2$
$v_f^2 = v_0^2 + 2 a \Delta s$	$\omega_f^2 = \omega_0^2 + 2\alpha \Delta\theta$
$\Delta s = \frac{v_0 + v_f}{2} t$	$\Delta\theta = \frac{\omega_0 + \omega_f}{2} t$

Yoom akka itti fayyadamtu

Gaaffilee rotational motion kan angular acceleration dhaabbataa qaban kinematics sararaawaa wajjin walbira qabuun furuuf fayyadamaa.

Of eeggadhu

Foormulaa keessa galchuun dura angular displacement gara radians jijjiiramuu qaba.

Grade 12 Physics, fuula 25-68 irraa qophaa'e; osoo hin ba'in dura yeroo lama mirkanaa'e

Torque and Angular AccelerationFoormulaa

\tau = I \alpha

\tau = I \alpha

$\tau$: Toorkii cuunfaa $N\,m$
$I$: Moomantii inarshiyaa $kg\,m^2$
$\alpha$: Saffisiisa kofaa $rad\,s^{-2}$

Yoom akka itti fayyadamtu

Net torque, moment of inertia fi angular acceleration qaama naanna'u tokkoo walqabsiisuuf fayyadamaa.

Of eeggadhu

Angular acceleration yeroo hunda safara radians per second squared ta'uu qaba.

Grade 12 Physics, fuula 25-68 irraa qophaa'e; osoo hin ba'in dura yeroo lama mirkanaa'e

Gaaffii shaakalaa boqonnaa 2

Gaaffii filannoo 10 boqonnaa kana irratti barreeffaman, tokkoon tokkoon isaanii deebii fi sababa qabu. Deebii banuu dura yaali. Gaaffiin kun Ingiliffaa tiin.

Gaaffii 1

In ideal projectile motion where air resistance is negligible, which of the following quantities remains constant throughout the flight?

A.Vertical velocity
B.Horizontal velocity
C.Vertical displacement
D.Total speed

Deebii fi sababa agarsiisi

Deebii: B Horizontal velocity

Sababa

Gravity acts only in the vertical direction. Because there is no horizontal force acting on the projectile, the horizontal acceleration is zero and the horizontal velocity v_x remains constant.

Gaaffii 2

At what launch angle with respect to the horizontal is the horizontal range of an inclined projectile maximum on level ground?

A.30 degrees
B.45 degrees
C.60 degrees
D.90 degrees

Deebii fi sababa agarsiisi

Deebii: B 45 degrees

Sababa

The range formula is R = (v₀² sin 2θ) / g. The factor sin 2θ reaches its maximum value of 1 when 2θ = 90°, which gives θ = 45°.

Gaaffii 3

What is the SI unit of moment of inertia (I)?

A.kg m
B.kg m^2
C.N m
D.rad/s^2

Deebii fi sababa agarsiisi

Deebii: B kg m^2

Sababa

The moment of inertia for a point mass is given by I = m r². The SI unit of mass is kg and distance is m, giving the unit kg m².

Gaaffii 4

According to Kepler's first law of planetary motion, what is the shape of a planet's orbit around the Sun?

A.A circle with the Sun at the center
B.A parabola with the Sun at the focus
C.An ellipse with the Sun at one focus
D.A hyperbola with the Sun at the center

Deebii fi sababa agarsiisi

Deebii: C An ellipse with the Sun at one focus

Sababa

Kepler's first law, also known as the law of ellipses, states that the orbit of a planet around the Sun is an ellipse with the Sun located at one of the two foci.

Gaaffii 5

An airplane flying horizontally at 50 m/s over Dire Dawa drops a food relief package from a height of 180 m. Taking g = 10 m/s², how long does it take for the package to hit the ground?

A.6.0 s
B.18.0 s
C.36.0 s
D.3.6 s

Deebii fi sababa agarsiisi

Deebii: A 6.0 s

Sababa

The initial vertical velocity is zero. Using Δy = (1/2) g t², we have 180 = (1/2)(10) t², which gives 5 t² = 180, so t² = 36 and t = 6.0 s.

Gaaffii 6

Two football players in Addis Ababa kick identical balls with the same initial speed v₀. Ball A is launched at an angle of 37° and Ball B is launched at 53°. Neglecting air resistance, which statement is true about their horizontal ranges on level ground?

A.Ball A travels a greater horizontal range.
B.Ball B travels a greater horizontal range.
C.Both balls achieve the same horizontal range.
D.The horizontal ranges cannot be compared without knowing the mass.

Deebii fi sababa agarsiisi

Deebii: C Both balls achieve the same horizontal range.

Sababa

Projectiles launched at complementary angles (angles that add up to 90°, such as 37° and 53°) with the same initial speed have the same horizontal range because sin(2 × 37°) = sin 74° and sin(2 × 53°) = sin 106° = sin 74°.

Gaaffii 7

A bicycle wheel in Hawassa accelerates uniformly from rest to an angular speed of 20 rad/s in 4.0 s. What is the angular acceleration of the wheel?

A.80 rad/s^2
B.5 rad/s^2
C.2.5 rad/s^2
D.0.2 rad/s^2

Deebii fi sababa agarsiisi

Deebii: B 5 rad/s^2

Sababa

Angular acceleration is defined as α = (ω_f - ω₀) / Δt. Substituting the given values: α = (20 rad/s - 0) / 4.0 s = 5 rad/s².

Gaaffii 8

A technician applies a force of 150 N perpendicular to a beam at a distance of 0.4 m from the pivot point. What is the magnitude of the torque produced?

A.375 N m
B.60 N m
C.150 N m
D.37.5 N m

Deebii fi sababa agarsiisi

Deebii: B 60 N m

Sababa

Torque is calculated using τ = r F sin θ. Since the force is perpendicular, θ = 90° and sin 90° = 1. Thus, τ = 0.4 m × 150 N × 1 = 60 N m.

Gaaffii 9

A projectile is launched from level ground in such a way that its horizontal range R is equal to 3 times its maximum height H. What is the launch angle θ?

A.30 degrees
B.37 degrees
C.45 degrees
D.53 degrees

Deebii fi sababa agarsiisi

Deebii: D 53 degrees

Sababa

The textbook gives the relation H = (R tan θ) / 4. Setting R = 3H gives H = (3H tan θ) / 4, so 1 = (3 tan θ) / 4, which leads to tan θ = 4/3 ≈ 1.33. Therefore, θ = 53°.

Gaaffii 10

A wheel with a moment of inertia of 2 kg m² changes its angular speed from 3 rad/s to 8 rad/s under the action of a constant net torque of 50 N m. How long does the wheel take to make this change?

A.0.20 s
B.0.50 s
C.2.00 s
D.5.00 s

Deebii fi sababa agarsiisi

Deebii: A 0.20 s

Sababa

First, use τ = I α to find the angular acceleration: α = τ / I = 50 / 2 = 25 rad/s². Next, using ω_f = ω₀ + α t, we get t = (ω_f - ω₀) / α = (8 - 3) / 25 = 5 / 25 = 0.20 s.

Appiin boqonnaa kana irratti gaaffii 12 qaba; utuu hojjettuu ni sirreessa. Kan asii jiran 10 jalqabaa ti.

Fuula kitaabaa boqonnaa 2

Boqonnaan 2 Maxxansa Ministeera Barnootaa keessatti fuula 25 hanga fuula 68 ti.

Maxxansa Ministeera Barnootaa, Ingiliffaa, Fuula 186

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Yaadannoo gabaabaa Fiizikii kutaa 12 hunda

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