Given:
Stopping distance range is d = (65, 70) ft.
The stopping distance, d, obeys this formula.
d = v²/(2μg)
where
v = speed of the vehicle
μ = 0.8, coefficient of static friction under good road conditions
g = acceleration due to gravity, 32.2 ft/s²
Therefore
v = √(2*0.8*32.2*d) = 7.178√d
Test d = 65 ft.
v = 7.178√(65) = 57.87 ft/s = (57.87/88)*60 = 39.5 mph
Test d = 70 ft.
v = 7.178√(70) = 60.05 ft/s = 40.9 mph
To be safe, the lower speed of 39.5 mph is preferred.
Answer: 40 mph
Change in position of object = Displacment
Answer:
Initial kinetic energy will be 2608200 lbf-ft
Initial speed will be 240.7488 m/sec
Explanation:
We have given height h = 900 ft
Mass of the object m = 90 lbm
Acceleration due to gravity 
Initial kinetic energy will be equal to potential energy of the object
So kinetic energy = potential energy = mgh 
This energy is also equal to the initial kinetic energy
So 
So initial velocity will be 240.7488 m/sec
Answer:
If the DNA of the proteins changes, this is called a mutation.
Explanation: Hope it helps you :))))
Have a good day
