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 stationary front" .
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D.) Nearly all light that strikes an "Opaque" <span>object will be absorbed by the object.
Hope this helps!</span>
M = 40 Kg , g=9.8 m/s² , h = 2 m
PE = m g h
PE = (40) (9.8) (2)
PE = 784 J
KE = PE
½m v² = m g h
½ v² = g h
½ v² = (9.8) (2)
½ v² = 19.6
v² = 19.6×2
v² = 39.2
V = √39.2
V = 6.26 m/s
KE = ½mv²
KE = ½(40) (6.26)²
KE =783.8 J
