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(295 kg)((10.0 m)/2 - x) = (63 kg)(x) solve for x x = 4.12
Yes you indeed have permission to write 32.66666666666666666666666666666 as 1/3 of 98. Masses m and 2m are balanced because the m one is twice the distance from the pivot as the 2m one, the two producing equal torque. The rod is then only balanced by the
a. option B b. option B
Given: m=0.2 R=5.5 r=1 l = 0.5*mR^2 =(0.5*0.2*5.5^2) =3.025 a. a=g[1 - I/(mr^2 + I)] =9.8[1 - (0.5*0.2*5.5^2)/(0.2*1^2 + (0.5*0.2*5.5^2))] =0.61 m/s^2 b. T=g/[r^2/I + 1/m] =9.8/[1^2/(0.5*0.2*5.5^2) + 1/0.2] =1.84 N
It= Ih+6I(tr) It=(6)*(1/2)^2+6(1/3*(1.5)*(1/2)^2) =2.25 kg*m^2
Note: we need to do part b first b. The average angular velocity is 2*(2pi) rad / 7 s = 4*pi / 7 rad/s = 1.8 rad/s That is the average. The question did not say, but we have assume the acceleration was constant. So the velocity, at t=7 s, is 2*1.8 rad/s =
CM of the stone is 9 cm from the edge, for the stick, 40 cm from the end, both by symmetry. putting the origin at the center of the stone r = 49/(1 + 8.3/1.8) = 8.73 cm that is from the center of the stone, which is 89 cm from the left end d = 89-8.73 =
a. momentum of car (Pc) = mass*velocity momentum = (528kg)*(17m/s) = 8976 kg*m/s Note: we need to find part c first b. momentum of truck (Pt) = mass*velocity momentum = (1352kg)*(11m/s) = 14872 kg*m/s c. tan(θ) = (momentum truck/momentum car) tan(59°) =
a. velocity of COM=[mᵢvᵢ + mᵢᵢvᵢᵢ] / [mᵢ + mᵢᵢ] = [109(4.9) + 83(-3.5)] / (109+83) = 1.27 b. initial velocity of car 1 = vᵢ - velocity of COM = 4.9-1.27 = 3.63 c. final velocity of car 1 = -3.63 d. final velocity of car 1 GRF = final
a. Vw = 3.4 cos(39) = 2.64 m/s b. Vb = 3.4 cos(51) = 2.14 m/s c. mVo = 1.43 kg * 3.4 m/s = 4.86 kg*m/s d.E = 1/2 m v^2 = 1/2 (1.43) (3.4^2) =8.27 J
(a) 0 kg*m/s (b) (15.8 - 4.5 - 5.4)kg = 5.9 kg (c) 0 = [4.5*27.6*cos162º + 5.4*21.5*cos293º + 5.9*V*cosΘ3] kg·m/s for V in m/s horizontal component V*cosΘ3 = 72.8/5.9 m/s = 12.3 m/s (d) 0 = [4.5*27.6*sin162º + 5.4*21.5*sin293º + 5.9*V*sinΘ3]
(0.224 kg)(14.8 m/s)*cos(27°) =2.954 kg*m/s The change in momentum is -2*2.954 kg*m/s, so the magnitude of the change is 5.91 kg*m/s.
(a) magnitude of the initial momentum is (0.238 kg)(12.5 m/s) = 2.975 kg*m/s. (b) The horizontal component of the initial momentum is (0.238 kg)(12.5 m/s)*cos(31) = 2.550 kg*m/s. The change in momentum is -2*2.550 kg*m/s, so the magnitude of the change is
60 g is not a small force. a. F=ma Force=4.5*60*9.8 N = 2646 N b. Force*time=19 time= 19/2646 =0.00718 s
a. Initial gravitational potential energy = 3*67 kg*9.8 m/s^2*9.9 m = 19501 J Final GPE = 1*67 kg*9.8 m/s^2*19 m = 12475 J fraction of the mechanical energy = 12475 J / 19501 J = 64/100 = 0.64 b. 3 jumper,s initial gravitational potential energy = final
Initial KE = 0.5*0.29*0.95^2 = 0.1308625 Final KE= (0.5*0.29*0.95^2)-(0.5*(0.29+0.55)*((0.29*0.95)/(0.29+0.55))^2) = 0.0451787 lost in the collision= Final KE - Initial KE = 0.0451787-0.1308625 = -0.08568 J -------> Answer
40
width = 11 lenght = 18
