Answer:
Explanation:
Let the velocity after first collision be v₁ and v₂ of car A and B . car A will bounce back .
velocity of approach = 1.5 - 0 = 1.5
velocity of separation = v₁ + v₂
coefficient of restitution = velocity of separation / velocity of approach
.8 = v₁ + v₂ / 1.5
v₁ + v₂ = 1.2
applying law of conservation of momentum
m x 1.5 + 0 = mv₂ - mv₁
1.5 = v₂ - v₁
adding two equation
2 v ₂= 2.7
v₂ = 1.35 m /s
v₁ = - .15 m / s
During second collision , B will collide with stationary A . Same process will apply in this case also. Let velocity of B and A after collision be v₃ and v₄.
For second collision ,
coefficient of restitution = velocity of separation / velocity of approach
.5 = v₃ + v₄ / 1.35
v₃ + v₄ = .675
applying law of conservation of momentum
m x 1.35 + 0 = mv₄ - mv₃
1.35 = v₄ - v₃
adding two equation
2 v ₄= 2.025
v₄ = 1.0125 m /s
v₃ = - 0 .3375 m / s
Which object? More information is needed to answer this question
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Answer:
1) F = 100N
2) a = 2 m/s²
3) m = 25 kg
Explanation:
1) F = ma ( F = ?, m = 10 kg, a = 10 m/s² )
F = 10×10
F = 100 N
2) F = ma ( F = 20N, m = 10 kg, a = ? )
20 = 10×a
10a = 20
a = 20/10
a = 2 m/s²
3)F = ma ( F = 100N, m = ?, a = 4 m/s² )
100 = m×4
4m = 100
m = 100/4
m = 25 kg
Hope that helps! Good luck!
Answer:
The force exerted by the rope on her arms is 273.7 N = 0.274 kN
Explanation:
Step 1: Data given
Mass of the ice skater = 55.6 kg
Velocity = 1.73 m/s
She then moves in a circle of radius 0.608 m around the pole.
Step 2:
Force exterted by the horizontal rope is the centripetal force acting on theice skater:
Fc = M*ac
⇒ with ac = v²/r
Fc = M * v²/r
Fc = 55.6 * 1.73²/0.608
Fc =273.69 N
The force exerted by the rope on her arms is 273.7 N = 0.274 kN
