Capitulo 4 Resolução- Moyses - Física 1

1) \n|F1| = |F2| \n 2 \n[\n\0451 2 \natan(f)\n2 \n\n\n\n\n\n\n\n|F| \n|x| = |F|\n|F| \n|F|\n|F|\n\n| F1 | = F2. 9 \n\n2\nF1 = 5.000,0 N\\ = [1, 0, 0] = F1 \n0\n|F2| = 2 1\n\n|F| = |F| = [|](5x) \t [0]\n\n[ h0 | 1 = 1 \n\na | F \n\n\t\n\t\n|F|\n\t\n\t F1 \nf \n|F1| = \n\nF = \\ c (3D \nf \n |F| = |\n\n --(Foy)\n\n2) \n\n|F| = 580m + 7} \n\n, F1(2F2 2)\n\00|, P = F2, c F2 = -2 = 0 \nF2 = x 0 = [5, 580] 580N\ny= 0.289\\\\\n|\n \n[ F2(830](3 \n\n\na2 (0.5) )= (2F1 F1\n\n, 1000,7 - 1 = \nF2 = \nF3 \n\n[ 1011] 3) \na \nF4 = \n {1] \nL (4)\n \nF4 15N\n\nF = 900x60(100) X 100N F4 = \n)\n\nF8 = 20\n\n 5) [ |F9 =0|] =\nF =0 [185 \n15kg, | m | \n778 [ d |F6 = | = ] = |F7) | \n| F|= F1\n9y\t] =F1- 20 = |F| =\n|F7|[ | = om [ | 1 2 → F6 | 5] \n\t\0. E\n0, 23 F = F| | F4 = 2.0 \nF5 = |F5| +2.5 + |F4F1 = 5| = 100\n\n = (17.4) hg + |[!----------------| \n\na 1 | \n 0 = (T )g\na = 0 100 \n300kg, \nF2 = F2 = [40°(8\t[ 330 906 = \n1/N(F3 [ \\(40° |=\n\n|] g\t = \n180[|1 x,\nF2] 30 F2 = (2003)\n] = G 10 + \n|1 (1.2) \n\n\t=T_ 3 08\\F9|\n|F8| =60\n|\n\ntan\nF3 = [( 7) \n0,5=0)+5M/s \n\x (10^4) \nF = -c \n where F | 500 + = 1.000x + 0.0 \n 98] X G | [Size \n|gx\nis =(1 —\n| \n|F21200 \n|S \n\n)\n\n Q[ |F2| | =1 a = 2 1 = 0\t, 5 0,5 = 3 = 3\n | = 10 F[p + (0 02)}} \n 1.2 | F) \n F| = {F/F -\\[ \n1(8.0 = += [ 1\n=Quest Oo) \nF = |3 |_ \nP = | |} = = |F41, 4 \n= o-5\nd = L [|F=F] (tm) \n= [F + \na = a=|0.799=0,01\tag |2,1000=2 01→ \n50N\ni = 200m\\t \n\] | [0][\n= 0111) -> F] = 1 \n|F|=1,2 = \n=[1.4 F3= \n, |020Q|)\n= 20\n\t\t2[| 2.5 )] \n[20.000|) kg\ng, \n1.00N | 10) 20° gradient shown R = -√2g h\n h = 3.20 sin(θ) = 2.45 m\n\n R (10/10 sec)\n m = 7.165; a = 5.162 m/s²\n\ny̅ = v₀t + (1/2)at² = (5.00) (3) + (1/2)(-9.81)(3)²\n= 7.25 + 5.5*2 - 0 = 0.95\n\nx̅ = ca = (a)(g) - 2 ²\n\na) R = mg\nb) T = mg - ma\nc) T = ma\n\n11) T = (M - m) a = (mg/M)\nT = F/M\n\n12) m\nF = ma = m(g - m/g)\n\n13) a) P = mg\n T = mgsin(θ) + T = -mg\n\nA: R² = (d + 2m/4)→\ng = R - ma = (g sin(θ)/g)\nF = ma = môm + 1m - sin θ +\n2 - 1.7mg\n\nb) T = mg/√(2g)\n