The aorta carries blood away from the heart at a speed of about 42 cm/s and has a diameter of approximately 1.1 cm. The aorta br
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Answer:
The minimum speed must the car must be 13.13 m/s.
Explanation:
The radius of the loop is 17.6 m. We need to find the minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top.
We know that, mg be the weight of car and rider, which is equal to the centripetal force.
So, the minimum speed must the car must be 13.13 m/s.
Answer:
a) v₂ = 4.2 m/s
b) v₂ = 5 m/s
Explanation:
a)
We will use the law of conservation of momentum here:
where,
m₁ = m₂ = mass of bowling pin = 1.8 kg
u₁ = speed of first pin before collsion = 5 m/s
u₂ = speed of second pin before collsion = 0 m/s
v₁ = speed of first pin after collsion = 0.8 m/s
v₂ = speed of second after before collsion = ?
Therefore,
<u>v₂ = 4.2 m/s</u>
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b)
We will use the law of conservation of momentum here:
where,
m₁ = m₂ = mass of bowling pin = 1.8 kg
u₁ = speed of first pin before collsion = 5 m/s
u₂ = speed of second pin before collsion = 0 m/s
v₁ = speed of first pin after collsion = 0 m/s
v₂ = speed of second after before collsion = ?
Therefore,
<u>v₂ = 5 m/s</u>