Question

How to solve ½ m1v12 + ½ m2v22 = ½ m1(v,1)2+½ m2(v,2)2 (energy)
(.5)( 0.5774)(0.03)^2 +( .5)(.308)( -0.015)^2 =( .5)(0.03)(0.89 - 2v2,)^2 + (.5)(-0.015)(v2,)^2

the 2 in front of the v means 2v and the 2 after is a subscript and the "," is a prime

Answer 1

Explanation:

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Answer 2

Answer:

½ m1v12 + ½ m2v22 = ½ m1(v,1)2+½ m2(v,2)2 (energy)

(.5)( 0.5774)(0.03)^2 +( .5)(.308)( -0.015)^2 =( .5)(0.03)(0.89 - 2v2,)^2 + (.5)(-0.015)(v2,)^2

Explanation:

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