Mtotal Vo MAVA + MBVB + McVc = use unit vectors to break velocities into components TA/D=TA-TD={-1.5,0,0}-{0,0, 5} {Vax,

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Mtotal Vo Mava Mbvb Mcvc Use Unit Vectors To Break Velocities Into Components Ta D Ta Td 1 5 0 0 0 0 5 Vax 1 (39.95 KiB) Viewed 36 times

Va = 394.64 m/s
Vb = 306.07m/s
Vc = 762.97m/s

Mtotal Vo Mava Mbvb Mcvc Use Unit Vectors To Break Velocities Into Components Ta D Ta Td 1 5 0 0 0 0 5 Vax 2 (39.95 KiB) Viewed 36 times

Mtotal Vo Mava Mbvb Mcvc Use Unit Vectors To Break Velocities Into Components Ta D Ta Td 1 5 0 0 0 0 5 Vax 3 (39.95 KiB) Viewed 36 times

Mtotal Vo Mava Mbvb Mcvc Use Unit Vectors To Break Velocities Into Components Ta D Ta Td 1 5 0 0 0 0 5 Vax 4 (48.71 KiB) Viewed 36 times

Mtotal Vo MAVA + MBVB + McVc = use unit vectors to break velocities into components TA/D=TA-TD={-1.5,0,0}-{0,0, 5} {Vax, Vay, Vaz} = VA {-1.5,0,-5) V1.5² +5³ 1.5m 5m 4.5m 2.5m 3m
An 8-kg shell moving with a velocity vo-(13.-11. -345) m/s explodes at Point D into three fragments A, B, and C of mass, respectively, 4 kg, 3 kg. and 1 kg. Knowing that the fragments hit the vertical wall at the points indicated, determine the speed of fragment A immediately after the explosion. Assume that elevation changes due to gravity may be neglected. Given: . . Initial mass 8 kg. initial velocity vo (13, -11, -345) m/s, explosion at - W Find: position (0, 0, 5) mass A 5m W 4 kg, final position of A=(-1.5,0,0) mass B 3 kg. final position of B= (4.5, 2.5, 0) - mass C 1 kg. final position of C = (0,-3,0) m Your Answer: 3m the speed of fragment A immediately after the explosion, (m/s) Answer Question 5 (1 point)