Fundamentos de Física - Vol 1 - 4 Ed - Resolução - Halliday Resnick

~ d = \sqrt{d_x^2 + d_y^2} = \sqrt{(-7.3)^2 + (3.3)^2} = 8.01 \approx 8. \theta_d = \frac{d_y}{d_x} = \frac{3.3}{-7.3} = 24.3^o. 90^o - 24.3^o = 65.7^o. a\ b 5 a + b\ b - a a + b\ b - a 35^o 400\ 40^o 860 \square a \triangle b O A\ B OB = |\overrightarrow{OB}|(- \theta i + \cos \theta j) = (860)(- 73^o i + \cos 73^o j) = - 822.42 i + 251.44 j \overrightarrow{OA} = |\overrightarrow{OA}|(\cos 40^o i + 40^o j) = (400)(\cos 40^o i + 40^o j) = 306.42 i + 257.11 j AB = \overrightarrow{OB} - \overrightarrow{OA}. = (-822.42 - 306.42, 251.44 - 257.11) = (-1128.84, -5.67), a_z = 5, a_y = 0, b_x = -4\ 35^o = -2.29, b_y = 4 \cos 35^o = 3.27. s = a + b s = (a_x + b_x, a_y + b_y) = (5 - 2.29, 0 + 3.27) \approx (2.7, 3.3), s = \sqrt{s_x^2 + s_y^2} = \sqrt{(2.7)^2 + (3.3)^2} = 4.26 \approx 4.2. \theta_s = \frac{s_y}{s_x} = \frac{3.27}{2.7} = 50.4^o \approx 50^o. 90^o - 50^o = 40^o d = b - a d = (-2.29 - 5, 3.27 - 0) \approx (-7.3, 3.3), ~ |AB| = \sqrt{(-1128.84)^2 + (-5.67)^2} = 1128.854\ \approx \ 1130\ . \left( \frac{-5.67}{-1128.84} \right) = 0.005 = 0.28^o, AB ~ ~ ~ t = 0.8757 \; t = -2.1206 \[12pt] \displaystyle v_{0y} = 6.1 + 9.8 \, t \[12pt] \displaystyle v_{0v} = 14.68 \simeq 14.7 \[12pt]~\[12pt] \displaystyle y_m = \dfrac{v_{0y}^2}{2g} = \dfrac{(14.7)^2}{2(9.8)} \; = \; 11 \; . \[12pt]~\[12pt] \displaystyle R \, = \, \dfrac{v_{0x} (2 v_{0y} )}{g} \, = \, \dfrac{(7.6) (2(14.7))}{9.8} \; = \; 22.8 \simeq 23 \; . \[12pt] \displaystyle v \, = \, \sqrt{v_{0x}^2 \, + v_{0y}^2} \[12pt]~\[12pt] \qquad \; = \sqrt{(7.6)^2 \,+ \,(14.7)^2} \; = \; 16.5 \simeq 17 \; . \[12pt]~\[12pt] \displaystyle \varphi \; = \; \text{-} \bigg( \dfrac{(v_{0x})}{v_{0x}} \bigg) \[12pt] \qquad \; \simeq \text{-} \bigg( \dfrac{\text{-}(14.7)}{7.6} \bigg) \; = \; 62.66^{\circ} \simeq 63^{\circ} \, , \[12pt] \; 63^{\circ} \[12pt] \; \def\strutG{ ule[10pt]{.3pt}{15pt}}\def\strutH{\rule[5pt]{.3pt}{15pt}} \begin{picture}(120,20)(0,-10) \begin{picture}(140,30)(-50,-50) \def\strutC{\rule[0pt]{1pt}{20pt}}\def\strutB{\rule[-5pt]{1pt}{30pt}} \begin{picture}(50,40)(-25,-35) \begin{picture}(40,40)(-5,-35) a \end{picture} \end{picture} \end{picture} \end{picture} ~ ---------------------------------~\[12pt] 9.1 \quad \quad \displaystyle \mathbf{v} \; = \; 7.6 \; i\; + \; 6.1 \; j \[12pt]~\[12pt] \vq \[12pt] \nu_{y}(t) \; = \; 6.1 \; = \; v_{0y} \text{-} g t , \[12pt] y(t) = 9.1 = v_{0} t - \dfrac{1}{2} gt^2 \[12pt]~\[12pt] v_{0v} \, 4.9 t^2 + 6.1 t - 9.1 = 0 \[12pt] \[ a P=mg=(630)(9.8)=6174 P=mg=(421)(9.8)=4125.8 a g=9.8 g=4.9 m=Pg=229.8=2.2 P=mg=(2.2)(4.9)=11 g=0 F_2=ma-F_1 F_1+F_2=ma F_1=20i a=-12 30⁰i-12cos30⁰j=-6i-10.4j F_2=(2)(-6)i+(2)(-10.4)j-20i=[-32i-21j] F_2=F^2_2x+F^2_2y=(−32)²+(−21)²=38 θ=F2yF2x=−21−32=0.656 33⁰F2x+180⁰=213⁰ mgmP=(11)(8.9)=108 213⁰ a16001.8500 T=mgθ=0N−mgcosθ=0 T=mg=(8.5)(9.8)30⁰=42 N=mgcosθ=(8.5)(9.8)cos30⁰=72 ama−mgθ=−mθ=−(9.8)30⁰=−4.9 a144107d=10−142v2=v20+2ad v=a v=1600a=444/1.8=247 1600/3.6=444 F=ma F=(500)(247)=1.2×105 F=ma d=v2−v202ad a=−(1.4×107)210(10−14)=−9.8×1027d a=Fma F=ma=(1.67×10−27)(9.8×1027)=16.4 1.67×10−27 θ=30⁰ y=12mv02yxv02= 12Fmxt2yv=12at2=12Fmt2 x=v0tx=30 t=x/v0 y=12Fmxv02=12(4.5×10−16/9.11×10−31)(30×10−3/1.2×107)=1.5×10−3=0.0015 a1.2×1074.5×10−169.11×10−31x8.5 aθ=37⁰Fmga=am F=mmF=52/40=0.13 x=15 θ≡37⁰T=mgF⁰at=Fmt=52/8.4=0.62 52 T−F=0,Tcosθ−mg=0. aT=mgθ=(3×10−4)(9.8)37⁰=2.21×10−3.T=mgcosθ=(3×10−4)(9.8)cos37⁰=3.68×10−330⁰TFF atxm=T12amt2 xt=x0−12at2. xm=xt 12amt2=x0−12at2t2=x0am+at xm=xmt2=x02xm=12amt2.x=xo xm=12amt2. tx0=2.6=(15)(0.13)0.13+0.62 a Fm1 kgNF1 mgF T of mg m m=x0.m1gm2 F2.1 −f m2 g 21153.2 3x10−4 cosT3. Glow in the dark FjP 1 m_2 g N_2 f -f \quad f m_1 F - f = m_1 a, v = 0 v_0 = -12 \quad y = -42 a = \frac { -v_0^2} {2y} = \frac { -(-12)^2}{2(-42)} = \frac {12}{7} = 1.71 \quad ^2, T = m(g + a) = (1600)(9.8 + 1.71) = 1.8 \times 10^4, a a e a \frac{ f}{ m_2} f = m_2 g. \quad f = 3.3 \times 10^5. a = \frac { F m_2 }{ m_1 + m_2 } \quad \frac { (3.2)(1.2)}{ 2.3 + 1.2} = 1.1. \quad \blacksquare F \quad \quad f = \frac { F m_1 }{ m_1 + m_2 } \quad \frac { (3.2)(2.3)}{ 2.3 + 1.2} = 2.1. F \quad\quad m_2\quad m_1 \quad \quad m = (80 + 5) = 85 mg - F_a = ma, \quad \qquad\qquad\quad a \qquad \qquad \qquad F_a = m(g-a) = (85)(9.8 - 2.5) = 620 F_a = m(g-a) = (85)(9.8 - 2.5) = 620 1600 F_p = m_p g + F_p - F_a = m_p a \quad F_p = m_p (a - g) + 620 = 580. m_p \quad F_p \quad \quad \quad \qquad \qquad \qquad \qquad \qquad \qquad \qquad m_p \quad F_p m_p a - \qquad \quad \qquad \quad (a-g) 42 \underline{ a } T \quad ma \quad \qquad \qquad \qquad m_p \quad m g = \qquad \qquad \qquad\qquad \qquad \quad \qquad \quad \qquad \underline{ a } - T \quad - \quad mg = 3260 \frac {mg}{ v^2 } \quad mg \quad T \quad m g \quad - mg \quad \underline{ a } \quad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 3260 2200^2 \quad \quad \quad \quad \quad 0.39 \frac { 39^2 }{ 39} v^2 = v_0^2 + 2ay. - \qquad \approx