The vertebral region is _________ to the scapular region.
1 answer:
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Answer:
Explanation:
The stunt will likely sustain serious injury in case of concrete blocks because the average force acting on the person will be more because concrete blocks do not squeeze to provide more time for the force to act on the body instead it acts for a small amount of interval.
As impulse is constant so time requires to act force on the body is more as compared to concrete block and thus average force in mattress case is less.
Answer:
(a) The maximum height achieved by the first ball, m₁ is 0.11 m
(a) The maximum height achieved by the second ball, m₂ ball is 0.44 m
Explanation:
Given;
mass of the first ball, m₁ = 1 kg
mass of the second ball, m₂ = 2 kg
The velocity of the first when released from a height of 1 m before collision;
u₁² = u₀² + 2gh
u₀ = 0, since it was released from rest
u₁² = 2gh
u₁² = 2 x 9.8 x 1
u₁² = 19.6
u₁ = √19.6
u₁ = 4.427 m/s
The velocity of the second ball before collision, u₂ = 0
Apply the principle of conservation of linear momentum, to determine the velocity of the balls after an elastic collision.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
where;
v₁ is the final velocity of the first ball after an elastic collision
v₂ is the final velocity of the second ball after an elastic collision
m₁u₁ + m₂(0) = m₁v₁ + m₂v₂
m₁u₁ = m₁v₁ + m₂v₂
1 x 4.427 = v₁ + 2v₂
v₁ + 2v₂ = 4.427
v₁ = 4.427 - 2v₂ ----- equation (1)
one directional velocity;
u₁ + v₁ = u₂ + v₂
u₂ = 0
u₁ + v₁ = v₂
v₁ = v₂ - u₁
v₁ = v₂ - 4.427 ------ equation (2)
Substitute v₁ into equation (1)
v₂ - 4.427 = 4.427 - 2v₂
3v₂ = 4.427 + 4.427
3v₂ = 8.854
v₂ = 8.854 / 3
v₂ = 2.95 m/s (→ forward direction)
v₁ = v₂ - 4.427
v₁ = 2.95 - 4.427
v₁ = - 1.477 m/s
v₁ = 1.477 m/s ( ← backward direction)
Apply the law of conservation of mechanical energy
(a) The maximum height achieved by the first ball (v₁ = 1.477 m/s)
(b) The maximum height achieved by the second ball (v₂ = 2.95 m/s)