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way performance, an 95.6 kg actor swings from a 4.30 m long cable that is horizontal when he starts. At the bottom of his arc,...Asked by bob
In a Broadway performance, an 80 kg actor swings from a 3.75 m long cable that is horizontal when he starts.?
At the bottom of his arc, he picks up his 55.0 kg costar in an inelastic collision. What maximum height do they reach after their upward swing?
At the bottom of his arc, he picks up his 55.0 kg costar in an inelastic collision. What maximum height do they reach after their upward swing?
Answers
Answered by
Elena
(a)
PE=mgh=mgL
KE = mv²/2
Law of conservation of energy
PE =KE
mgL= mv²/2
v=sqrt(2gL)=...
(b) Law of conservation of linear momentum
mv= (m+m1)u
u= m•v/(m+m1)=...
(c)
Law of conservation of energy
KE1=PE1
(m+m1) •u²/2 =(m+m1) •g•h1
h1=(m+m1) •u²/2 • (m+m1) •g =...
PE=mgh=mgL
KE = mv²/2
Law of conservation of energy
PE =KE
mgL= mv²/2
v=sqrt(2gL)=...
(b) Law of conservation of linear momentum
mv= (m+m1)u
u= m•v/(m+m1)=...
(c)
Law of conservation of energy
KE1=PE1
(m+m1) •u²/2 =(m+m1) •g•h1
h1=(m+m1) •u²/2 • (m+m1) •g =...
Answered by
Kel
KE1=PE1
in the end should actually be...
h1=(m+m1)(u^2)/2*g*(m+m1)
where the sum of the masses cancel and you're left with: (u^2)/2*g
in the end should actually be...
h1=(m+m1)(u^2)/2*g*(m+m1)
where the sum of the masses cancel and you're left with: (u^2)/2*g
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