2 3 5 6 30 3 Motal Vo MAVA+MBVB+ McVe T use unit vectors to break velocities into components TA/D=TA-TD={-1.5,0,0}-{0,0,

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2 3 5 6 30 3 Motal Vo MAVA+MBVB+ McVe T use unit vectors to break velocities into components TA/D=TA-TD={-1.5,0,0}-{0,0,

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2 3 5 6 30 3 Motal Vo Mava Mbvb Mcve T Use Unit Vectors To Break Velocities Into Components Ta D Ta Td 1 5 0 0 0 0 1
2 3 5 6 30 3 Motal Vo Mava Mbvb Mcve T Use Unit Vectors To Break Velocities Into Components Ta D Ta Td 1 5 0 0 0 0 1 (25.09 KiB) Viewed 41 times
2 3 5 6 30 3 Motal Vo Mava Mbvb Mcve T Use Unit Vectors To Break Velocities Into Components Ta D Ta Td 1 5 0 0 0 0 2
2 3 5 6 30 3 Motal Vo Mava Mbvb Mcve T Use Unit Vectors To Break Velocities Into Components Ta D Ta Td 1 5 0 0 0 0 2 (29.12 KiB) Viewed 41 times
2 3 5 6 30 3 Motal Vo MAVA+MBVB+ McVe T use unit vectors to break velocities into components TA/D=TA-TD={-1.5,0,0}-{0,0, 5} {Vaz, Vay, Vaz} = V₁ (-150,-5) 5m 4.5m 2.5m 3m
5m An 8-kg shell moving with a velocity vo-[15. -11.-350) 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: Find: 3m . Initial mass 8 kg, initial velocity vo - (15.-11. -350) m/s, explosion at position (0, 0, 5) mass A 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) Your Answer: the speed of fragment C immediately after the explosion (m/s).
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