Wasif

This page lists questions and answers that were posted by visitors named Wasif.

Questions

The following questions were asked by visitors named Wasif.

Answers

The following answers were posted by visitors named Wasif.

width = 11 lenght = 18

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

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 k...

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) 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...

(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) 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] kg·m/s for V...

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. 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 velocity of car 1 + v...

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°) = (...

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 = 8...

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 =...

It= Ih+6I(tr) It=(6)*(1/2)^2+6(1/3*(1.5)*(1/2)^2) =2.25 kg*m^2

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

a. option B b. option B

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...

(295 kg)((10.0 m)/2 - x) = (63 kg)(x) solve for x x = 4.12