Bob Beamon's 1968 Olympic long jump set a world record which remains unbroken to this day. This amazing jump resulted from an in
2 answers:
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PART a)
Before Drew throw Lily in forwards direction they both stays at rest
So initial speed of both of them is zero
So here we can say that initial momentum of both of them is zero
So total momentum of the system initially = ZERO
PART b)
Since there is no external force on the system of two
so there will be no change in the momentum of this system and it will remain same as initial momentum
So final momentum of both of them will be ZERO
PART c)
As we know that momentum of both will be zero always
so we have
in opposite direction
Refer to the diagram shown below.
The hoist is in static equilibrium supported by tensions in the two ropes.
For horizontal force balance, obtain
T₃ cos 50 = T₂ cos 38
0.6428T₃ = 0.788T₂
T₃ = 1.2259T₂ (1)
For vertical force balance, obtain
T₂ sin 38 + T₃ sin 50 = 350
0.6157T₂ + 0.766T₃ = 350 (2)
Substitute (1) into (2).
0.6157T₂ + 0.766(1.2259T₂) = 350
1.5547T₂ = 350
T₂ = 225.124 N
T₃ = 1.2259(225.124) = 275.979
Answer:
T₂ = 225.12 N
T₃ = 275.98 N
The hoist is in static equilibrium supported by tensions in the two ropes.
For horizontal force balance, obtain
T₃ cos 50 = T₂ cos 38
0.6428T₃ = 0.788T₂
T₃ = 1.2259T₂ (1)
For vertical force balance, obtain
T₂ sin 38 + T₃ sin 50 = 350
0.6157T₂ + 0.766T₃ = 350 (2)
Substitute (1) into (2).
0.6157T₂ + 0.766(1.2259T₂) = 350
1.5547T₂ = 350
T₂ = 225.124 N
T₃ = 1.2259(225.124) = 275.979
Answer:
T₂ = 225.12 N
T₃ = 275.98 N