1.1 Vektor Algebra — Masalalar

Jami: 25 ta | Yechim bilan: ✅

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📊 Qiyinlik Darajalari

• ⭐⭐ Asosiy (1-8)
• ⭐⭐⭐ O'rtacha (9-17)
• ⭐⭐⭐⭐ Murakkab (18-25)

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Asosiy Masalalar (1-8)

Masala 1 ⭐⭐

$\vec{a} = (3, -2, 5)$ va $\vec{b} = (-1, 4, 2)$ vektorlar berilgan.

Toping: a) $\vec{a} + \vec{b}$ b) $\vec{a} - \vec{b}$ c) $2\vec{a} + 3\vec{b}$

Yechim

a) $\vec{a} + \vec{b} = (3-1, -2+4, 5+2) = (2, 2, 7)$

b) $\vec{a} - \vec{b} = (3-(-1), -2-4, 5-2) = (4, -6, 3)$

c) $2\vec{a} + 3\vec{b} = 2(3, -2, 5) + 3(-1, 4, 2)$ $= (6, -4, 10) + (-3, 12, 6) = (3, 8, 16)$

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Masala 2 ⭐⭐

$\vec{v} = (4, 3, 0)$ vektorning uzunligini va birlik vektorini toping.

Yechim

Uzunlik: $|\vec{v}| = \sqrt{4^2 + 3^2 + 0^2} = \sqrt{16 + 9} = \sqrt{25} = 5$

Birlik vektor: $\hat{v} = \frac{\vec{v}}{|\vec{v}|} = \frac{1}{5}(4, 3, 0) = (0.8, 0.6, 0)$

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Masala 3 ⭐⭐

$\vec{a} = (2, 1, -1)$ va $\vec{b} = (1, -1, 2)$ uchun skalyar ko'paytmani hisoblang.

Yechim

$\vec{a} \cdot \vec{b} = 2(1) + 1(-1) + (-1)(2) = 2 - 1 - 2 = -1$

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Masala 4 ⭐⭐

$\vec{a} = (1, 2, 3)$ va $\vec{b} = (4, 5, 6)$ vektorlar orasidagi burchakni toping.

Yechim

$\vec{a} \cdot \vec{b} = 1(4) + 2(5) + 3(6) = 4 + 10 + 18 = 32$

$|\vec{a}| = \sqrt{1 + 4 + 9} = \sqrt{14}$

$|\vec{b}| = \sqrt{16 + 25 + 36} = \sqrt{77}$

$\cos\theta = \frac{32}{\sqrt{14} \cdot \sqrt{77}} = \frac{32}{\sqrt{1078}} \approx 0.974$

$\theta = \arccos(0.974) \approx 12.9°$

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Masala 5 ⭐⭐

$\vec{a} = (1, 2, 0)$ va $\vec{b} = (3, -1, 4)$ uchun vektor ko'paytmani hisoblang.

Yechim

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 0 \\ 3 & -1 & 4 \end{vmatrix}$

$= \hat{i}(2 \cdot 4 - 0 \cdot (-1)) - \hat{j}(1 \cdot 4 - 0 \cdot 3) + \hat{k}(1 \cdot (-1) - 2 \cdot 3)$

$= \hat{i}(8) - \hat{j}(4) + \hat{k}(-7)$

$\vec{a} \times \vec{b} = (8, -4, -7)$

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Masala 6 ⭐⭐

$A(1, 2, 3)$, $B(4, 5, 6)$, $C(2, 1, 0)$ nuqtalari berilgan. $\overrightarrow{AB}$ va $\overrightarrow{AC}$ vektorlarini toping.

Yechim

$\overrightarrow{AB} = B - A = (4-1, 5-2, 6-3) = (3, 3, 3)$

$\overrightarrow{AC} = C - A = (2-1, 1-2, 0-3) = (1, -1, -3)$

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Masala 7 ⭐⭐

$\vec{a} = (2, -1, 3)$ vektorning $\vec{b} = (1, 1, 1)$ ustiga proyeksiyasini toping.

Yechim

Skalyar proyeksiya: $\text{proj}_{\vec{b}}\vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{2(-1)+(-1)(1)+3(1)}{\sqrt{3}} = \frac{2-1+3}{\sqrt{3}} = \frac{4}{\sqrt{3}} \approx 2.31$

Vektor proyeksiya: $\text{proj}_{\vec{b}}\vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\vec{b} = \frac{4}{3}(1, 1, 1) = \left(\frac{4}{3}, \frac{4}{3}, \frac{4}{3}\right)$

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Masala 8 ⭐⭐

Quyidagi vektorlar perpendikularmi? $\vec{a} = (2, -3, 1)$ va $\vec{b} = (1, 1, 1)$

Yechim

$\vec{a} \cdot \vec{b} = 2(1) + (-3)(1) + 1(1) = 2 - 3 + 1 = 0$

Ha, $\vec{a} \cdot \vec{b} = 0$ bo'lgani uchun vektorlar perpendikular.

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O'rtacha Masalalar (9-17)

Masala 9 ⭐⭐⭐

$\vec{a} = (1, 2, -2)$ va $\vec{b} = (3, -6, 2)$ vektorlarga perpendikular bo'lgan birlik vektorni toping.

Yechim

Perpendikular vektor: $\vec{n} = \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -2 \\ 3 & -6 & 2 \end{vmatrix}$

$= \hat{i}(2 \cdot 2 - (-2)(-6)) - \hat{j}(1 \cdot 2 - (-2) \cdot 3) + \hat{k}(1(-6) - 2 \cdot 3)$

$= \hat{i}(4 - 12) - \hat{j}(2 + 6) + \hat{k}(-6 - 6)$

$\vec{n} = (-8, -8, -12)$

Uzunligi: $|\vec{n}| = \sqrt{64 + 64 + 144} = \sqrt{272} = 4\sqrt{17}$

Birlik vektor: $\hat{n} = \frac{1}{4\sqrt{17}}(-8, -8, -12) = \left(-\frac{2}{\sqrt{17}}, -\frac{2}{\sqrt{17}}, -\frac{3}{\sqrt{17}}\right)$

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Masala 10 ⭐⭐⭐

$A(1, 0, 0)$, $B(0, 1, 0)$, $C(0, 0, 1)$ nuqtalardan o'tuvchi uchburchak yuzini toping.

Yechim

$\overrightarrow{AB} = (-1, 1, 0)$ $\overrightarrow{AC} = (-1, 0, 1)$

$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{vmatrix}$

$= \hat{i}(1 \cdot 1 - 0 \cdot 0) - \hat{j}((-1) \cdot 1 - 0 \cdot (-1)) + \hat{k}((-1) \cdot 0 - 1 \cdot (-1))$

$= (1, 1, 1)$

$|\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{1 + 1 + 1} = \sqrt{3}$

$S = \frac{1}{2}\sqrt{3} \approx 0.866$

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Masala 11 ⭐⭐⭐

Robot qo'lining oxirgi nuqtasiga $\vec{F} = (10, 5, -3)$ N kuch ta'sir qilmoqda. Aylanish markazi $O(0, 0, 0)$, kuch qo'yilgan nuqta $P(0.3, 0.2, 0.5)$ m. Momentni hisoblang.

Yechim

$\vec{\tau} = \vec{r} \times \vec{F}$

$\vec{r} = (0.3, 0.2, 0.5)$ $\vec{F} = (10, 5, -3)$

$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0.3 & 0.2 & 0.5 \\ 10 & 5 & -3 \end{vmatrix}$

$= \hat{i}(0.2 \cdot (-3) - 0.5 \cdot 5) - \hat{j}(0.3 \cdot (-3) - 0.5 \cdot 10) + \hat{k}(0.3 \cdot 5 - 0.2 \cdot 10)$

$= \hat{i}(-0.6 - 2.5) - \hat{j}(-0.9 - 5) + \hat{k}(1.5 - 2)$

$\vec{\tau} = (-3.1, 5.9, -0.5) \text{ N·m}$

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Masala 12 ⭐⭐⭐

Dron $\vec{v} = (5, 3, 2)$ m/s tezlik bilan uchmoqda. Shamol $\vec{w} = (-2, 1, 0)$ m/s. Natijaviy tezlik va uning yerga nisbatan burchagini toping.

Yechim

Natijaviy tezlik: $\vec{v}_{net} = \vec{v} + \vec{w} = (5-2, 3+1, 2+0) = (3, 4, 2)$

Tezlik kattaligi: $|\vec{v}_{net}| = \sqrt{9 + 16 + 4} = \sqrt{29} \approx 5.39 \text{ m/s}$

Yer tekisligi (xy) bilan burchak: $\theta = \arctan\left(\frac{v_z}{\sqrt{v_x^2 + v_y^2}}\right) = \arctan\left(\frac{2}{\sqrt{9+16}}\right) = \arctan\left(\frac{2}{5}\right) \approx 21.8°$

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Masala 13 ⭐⭐⭐

$\vec{a}$, $\vec{b}$, $\vec{c}$ vektorlar berilgan: $\vec{a} = (1, 0, 1)$, $\vec{b} = (0, 1, 1)$, $\vec{c} = (1, 1, 0)$. Ular hosil qilgan parallelepiped hajmini toping.

Yechim

Aralash ko'paytma (hajm): $V = |\vec{a} \cdot (\vec{b} \times \vec{c})|$

Avval $\vec{b} \times \vec{c}$: $\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{vmatrix}$

$= \hat{i}(1 \cdot 0 - 1 \cdot 1) - \hat{j}(0 \cdot 0 - 1 \cdot 1) + \hat{k}(0 \cdot 1 - 1 \cdot 1)$

$= (-1, 1, -1)$

Keyin: $\vec{a} \cdot (\vec{b} \times \vec{c}) = (1, 0, 1) \cdot (-1, 1, -1) = -1 + 0 - 1 = -2$

$V = |-2| = 2 \text{ (birlik}^3)$

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Masala 14 ⭐⭐⭐

$\vec{a} = (2, 1, -1)$ vektorni $\vec{b} = (1, 0, 0)$ ga parallel va perpendikular komponentlarga ajrating.

Yechim

$\vec{b}$ ga parallel komponent (proyeksiya): $\vec{a}_{\parallel} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\vec{b} = \frac{2(1) + 1(0) + (-1)(0)}{1}(1, 0, 0) = 2(1, 0, 0) = (2, 0, 0)$

Perpendikular komponent: $\vec{a}_{\perp} = \vec{a} - \vec{a}_{\parallel} = (2, 1, -1) - (2, 0, 0) = (0, 1, -1)$

Tekshirish: $\vec{a}_{\parallel} + \vec{a}_{\perp} = (2, 0, 0) + (0, 1, -1) = (2, 1, -1) = \vec{a}$ ✓

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Masala 15 ⭐⭐⭐

Dekart koordinatalarida $(3, 4, 5)$ nuqtani silindr koordinatalarga o'tkazing.

Yechim

$\rho = \sqrt{x^2 + y^2} = \sqrt{9 + 16} = 5$

$\phi = \arctan\left(\frac{y}{x}\right) = \arctan\left(\frac{4}{3}\right) \approx 53.13° \approx 0.927 \text{ rad}$

$z = 5$

Javob: $(\rho, \phi, z) = (5, 53.13°, 5)$ yoki $(5, 0.927, 5)$ rad

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Masala 16 ⭐⭐⭐

Sferik koordinatalar $(r, \theta, \phi) = (10, 60°, 45°)$ ni Dekart koordinatalarga o'tkazing.

Yechim

$x = r\sin\theta\cos\phi = 10 \sin(60°)\cos(45°) = 10 \cdot \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2}$ $x = 10 \cdot \frac{\sqrt{6}}{4} \approx 6.12$

$y = r\sin\theta\sin\phi = 10 \sin(60°)\sin(45°) = 10 \cdot \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2}$ $y \approx 6.12$

$z = r\cos\theta = 10\cos(60°) = 10 \cdot 0.5 = 5$

Javob: $(x, y, z) \approx (6.12, 6.12, 5)$

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Masala 17 ⭐⭐⭐

$\vec{F}_1 = (100, 50, 0)$ N va $\vec{F}_2 = (-30, 80, 20)$ N kuchlar bitta nuqtaga ta'sir qilmoqda. Natijaviy kuchning kattaligi va yo'nalish kosinuslarini toping.

Yechim

Natijaviy kuch: $\vec{F} = \vec{F}_1 + \vec{F}_2 = (70, 130, 20) \text{ N}$

Kattalik: $|\vec{F}| = \sqrt{70^2 + 130^2 + 20^2} = \sqrt{4900 + 16900 + 400} = \sqrt{22200} \approx 149 \text{ N}$

Yo'nalish kosinuslari: $\cos\alpha = \frac{F_x}{|\vec{F}|} = \frac{70}{149} \approx 0.470$ $\cos\beta = \frac{F_y}{|\vec{F}|} = \frac{130}{149} \approx 0.872$ $\cos\gamma = \frac{F_z}{|\vec{F}|} = \frac{20}{149} \approx 0.134$

Tekshirish: $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 0.221 + 0.760 + 0.018 \approx 1$ ✓

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Murakkab Masalalar (18-25)

Masala 18 ⭐⭐⭐⭐

Quadcopter dronning 4 ta motori quyidagi itarish kuchlarini bermoqda (birlik vektorlar bilan):

• Motor 1: $F_1 = 5$ N, $\hat{n}_1 = (0, 0, 1)$
• Motor 2: $F_2 = 6$ N, $\hat{n}_2 = (0.1, 0, 0.995)$
• Motor 3: $F_3 = 5$ N, $\hat{n}_3 = (-0.1, 0.1, 0.990)$
• Motor 4: $F_4 = 4$ N, $\hat{n}_4 = (0, -0.1, 0.995)$

Natijaviy kuch vektorini va uning vertikal bilan burchagini toping.

Yechim

Kuch vektorlari: $\vec{F}_1 = 5(0, 0, 1) = (0, 0, 5)$ $\vec{F}_2 = 6(0.1, 0, 0.995) = (0.6, 0, 5.97)$ $\vec{F}_3 = 5(-0.1, 0.1, 0.990) = (-0.5, 0.5, 4.95)$ $\vec{F}_4 = 4(0, -0.1, 0.995) = (0, -0.4, 3.98)$

Natijaviy: $\vec{F}_{total} = (0.1, 0.1, 19.9) \text{ N}$

Kattalik: $|\vec{F}_{total}| = \sqrt{0.01 + 0.01 + 396.01} \approx 19.9 \text{ N}$

Vertikal bilan burchak: $\cos\theta = \frac{F_z}{|\vec{F}|} = \frac{19.9}{19.9} \approx 0.9999$ $\theta \approx 0.41°$

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Masala 19 ⭐⭐⭐⭐

Robot qo'lining uchta bo'g'ini bor. Har bir bo'g'inning uzunligi va burchagi:

• $L_1 = 0.5$ m, $\theta_1 = 30°$ (gorizontaldan)
• $L_2 = 0.3$ m, $\theta_2 = 45°$ (birinchidan)
• $L_3 = 0.2$ m, $\theta_3 = -20°$ (ikkinchidan)

Oxirgi nuqta (end effector) pozitsiyasini toping (2D tekislikda).

Yechim

Har bir bo'g'in pozitsiyasi (kümülatif burchak): $\phi_1 = 30°$ $\phi_2 = 30° + 45° = 75°$ $\phi_3 = 75° + (-20°) = 55°$

Pozitsiyalar: $x_1 = L_1\cos\phi_1 = 0.5\cos(30°) = 0.433 \text{ m}$ $y_1 = L_1\sin\phi_1 = 0.5\sin(30°) = 0.25 \text{ m}$

$x_2 = x_1 + L_2\cos\phi_2 = 0.433 + 0.3\cos(75°) = 0.433 + 0.078 = 0.511 \text{ m}$ $y_2 = y_1 + L_2\sin\phi_2 = 0.25 + 0.3\sin(75°) = 0.25 + 0.290 = 0.540 \text{ m}$

$x_3 = x_2 + L_3\cos\phi_3 = 0.511 + 0.2\cos(55°) = 0.511 + 0.115 = 0.626 \text{ m}$ $y_3 = y_2 + L_3\sin\phi_3 = 0.540 + 0.2\sin(55°) = 0.540 + 0.164 = 0.704 \text{ m}$

End Effector: $(0.626, 0.704)$ m

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Masala 20 ⭐⭐⭐⭐

Raketa traektoriyasi parametrik ko'rinishda berilgan: $\vec{r}(t) = (100t, 200t - 5t^2, 50t)$ bu yerda $t$ sekundlarda, koordinatalar metrlarda.

a) $t = 10$ s dagi tezlik vektorini toping b) $t = 10$ s dagi tezlanish vektorini toping c) Tezlik va tezlanish orasidagi burchakni toping

Yechim

a) Tezlik (hosilasi): $\vec{v}(t) = \frac{d\vec{r}}{dt} = (100, 200 - 10t, 50)$

$t = 10$ da: $\vec{v}(10) = (100, 200 - 100, 50) = (100, 100, 50) \text{ m/s}$

b) Tezlanish: $\vec{a}(t) = \frac{d\vec{v}}{dt} = (0, -10, 0) \text{ m/s}^2$

Bu doimiy (gravitatsiya).

c) Burchak: $\vec{v} \cdot \vec{a} = 100(0) + 100(-10) + 50(0) = -1000$

$|\vec{v}| = \sqrt{10000 + 10000 + 2500} = \sqrt{22500} = 150 \text{ m/s}$ $|\vec{a}| = 10 \text{ m/s}^2$

$\cos\theta = \frac{-1000}{150 \cdot 10} = -\frac{2}{3}$ $\theta = \arccos(-0.667) \approx 131.8°$

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Masala 21 ⭐⭐⭐⭐

Dron kamerasi yo'nalishini belgilovchi birlik vektor $\hat{c} = (0.6, 0.8, 0)$ (gorizontal). Kamerani vertikal bo'yicha $\alpha = 30°$ ga pastga burilishi kerak. Yangi yo'nalish vektorini toping.

Yechim

Vertikal pastga burilish uchun, gorizontal komponent saqlanib, vertikal komponent qo'shiladi.

Yangi vektor (normallashtirilmagan): $\vec{c}' = (\cos(30°) \cdot 0.6, \cos(30°) \cdot 0.8, -\sin(30°))$ $= (0.866 \cdot 0.6, 0.866 \cdot 0.8, -0.5)$ $= (0.520, 0.693, -0.5)$

Uzunligi: $|\vec{c}'| = \sqrt{0.270 + 0.480 + 0.25} = \sqrt{1.0} = 1$

Bu allaqachon birlik vektor.

Javob: $\hat{c}' = (0.520, 0.693, -0.5)$

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Masala 22 ⭐⭐⭐⭐

Uchta GPS sun'iy yo'ldoshi quyidagi pozitsiyalarda (ECEF koordinatalari, km):

• $S_1 = (20000, 15000, 10000)$
• $S_2 = (18000, 20000, 8000)$
• $S_3 = (22000, 12000, 12000)$

Agar qabul qiluvchi nuqta $R = (0, 0, 6371)$ (Yer sirtida) bo'lsa, har bir sun'iy yo'ldoshgacha masofani toping.

Yechim

$d_1 = |S_1 - R| = |(20000, 15000, 3629)|$ $= \sqrt{20000^2 + 15000^2 + 3629^2} = \sqrt{400M + 225M + 13.17M}$ $= \sqrt{638.17M} \approx 25262 \text{ km}$

$d_2 = |S_2 - R| = |(18000, 20000, 1629)|$ $= \sqrt{324M + 400M + 2.65M} = \sqrt{726.65M} \approx 26956 \text{ km}$

$d_3 = |S_3 - R| = |(22000, 12000, 5629)|$ $= \sqrt{484M + 144M + 31.69M} = \sqrt{659.69M} \approx 25684 \text{ km}$

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Masala 23 ⭐⭐⭐⭐

Propeller aylanish o'qi $\hat{n} = (0, 0, 1)$ (vertikal). Aylanish tezligi $\omega = 500$ rad/s. Propeller uchidagi nuqta pozitsiyasi $\vec{r} = (0.15, 0, 0)$ m (aylanish markazidan).

a) Nuqtaning chiziqli tezligini toping b) Markazga intilma tezlanishni toping

Yechim

a) Chiziqli tezlik: $\vec{v} = \vec{\omega} \times \vec{r}$

Bu yerda $\vec{\omega} = \omega\hat{n} = (0, 0, 500)$ rad/s

$\vec{v} = (0, 0, 500) \times (0.15, 0, 0)$ $= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & 500 \\ 0.15 & 0 & 0 \end{vmatrix}$ $= (0 \cdot 0 - 500 \cdot 0, 500 \cdot 0.15 - 0 \cdot 0, 0 \cdot 0 - 0 \cdot 0.15)$ $= (0, 75, 0) \text{ m/s}$

Tezlik kattaligi: $|\vec{v}| = 75$ m/s

b) Markazga intilma tezlanish: $a_c = \frac{v^2}{r} = \frac{75^2}{0.15} = \frac{5625}{0.15} = 37500 \text{ m/s}^2$

Yoki: $a_c = \omega^2 r = 500^2 \cdot 0.15 = 37500 \text{ m/s}^2$

Yo'nalish: markazga qarab, $\vec{a}_c = -37500\hat{r} = (-37500, 0, 0)$ m/s²

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Masala 24 ⭐⭐⭐⭐

Ikki robot qo'li koordinatsiyalangan harakat qilmoqda:

• Robot A: $\vec{r}_A(t) = (0.5 + 0.1t, 0.3 + 0.05t, 0.8 - 0.02t)$ m
• Robot B: $\vec{r}_B(t) = (0.7 - 0.08t, 0.2 + 0.06t, 0.6 + 0.03t)$ m

a) Qaysi vaqtda ular eng yaqin bo'ladi? b) Minimal masofani toping

Yechim

Orasidagi masofa vektori: $\vec{d}(t) = \vec{r}_B - \vec{r}_A = (0.2 - 0.18t, -0.1 + 0.01t, -0.2 + 0.05t)$

Masofa kvadrati: $d^2(t) = (0.2 - 0.18t)^2 + (-0.1 + 0.01t)^2 + (-0.2 + 0.05t)^2$

Minimumni topish uchun hosilani nolga tenglashtiramiz: $\frac{d(d^2)}{dt} = 2(0.2 - 0.18t)(-0.18) + 2(-0.1 + 0.01t)(0.01) + 2(-0.2 + 0.05t)(0.05) = 0$

$-0.18(0.2 - 0.18t) + 0.01(-0.1 + 0.01t) + 0.05(-0.2 + 0.05t) = 0$ $-0.036 + 0.0324t - 0.001 + 0.0001t - 0.01 + 0.0025t = 0$ $0.035t = 0.047$ $t = 1.34 \text{ s}$

b) Minimal masofa: $\vec{d}(1.34) = (0.2 - 0.241, -0.1 + 0.013, -0.2 + 0.067)$ $= (-0.041, -0.087, -0.133)$

$d_{min} = \sqrt{0.00168 + 0.00757 + 0.0177} \approx 0.164 \text{ m}$

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Masala 25 ⭐⭐⭐⭐

Raketa uchish paytida quyidagi kuchlar ta'sir qilmoqda:

• Itarish: $\vec{T} = (0, 0, 50000)$ N
• Gravitatsiya: $\vec{G} = (0, 0, -9810)$ N (massa 1000 kg)
• Havo qarshiligi: $\vec{D} = -0.1|\vec{v}|^2\hat{v}$, tezlik $\vec{v} = (10, 5, 200)$ m/s

a) Havo qarshiligi vektorini hisoblang b) Natijaviy kuchni toping c) Tezlanishni toping

Yechim

a) Havo qarshiligi: $|\vec{v}| = \sqrt{100 + 25 + 40000} = \sqrt{40125} \approx 200.3 \text{ m/s}$

$\hat{v} = \frac{\vec{v}}{|\vec{v}|} = \frac{(10, 5, 200)}{200.3} \approx (0.05, 0.025, 0.999)$

$\vec{D} = -0.1 \cdot 200.3^2 \cdot (0.05, 0.025, 0.999)$ $= -4012 \cdot (0.05, 0.025, 0.999)$ $= (-200.6, -100.3, -4008) \text{ N}$

b) Natijaviy kuch: $\vec{F} = \vec{T} + \vec{G} + \vec{D}$ $= (0, 0, 50000) + (0, 0, -9810) + (-200.6, -100.3, -4008)$ $= (-200.6, -100.3, 36182) \text{ N}$

c) Tezlanish: $\vec{a} = \frac{\vec{F}}{m} = \frac{(-200.6, -100.3, 36182)}{1000}$ $= (-0.2, -0.1, 36.2) \text{ m/s}^2$

Kattalik: $|\vec{a}| = \sqrt{0.04 + 0.01 + 1310.4} \approx 36.2 \text{ m/s}^2$

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✅ Tekshirish Ro'yxati

• 1-8: Asosiy operatsiyalar
• 9-17: O'rtacha masalalar
• 18-25: Murakkab masalalar

Maqsad: Kamida 20 ta masalani mustaqil yechish

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