If there are 272.5 rotations in a mile, and 5280 ft in a mile, then there are 5280 ft in 272.5 rotations. Divide 5280 by 272.5 gives us 19.4 ft/rotation.
That's the circumference of the wheel, and since c = πd, 19.4/π = d, d = 6.2 ft. The diameter of the wheel is how tall it is.
Does this help? Any questions?
It is 46 if you divide 34 by 16 you get 2.125 then divide 100 by 2.125 and you get 47.05 the 0 means go down if you are rounding which brings it to 46
Velocity=Distance/Time
Time(t)=Distance/Velocity
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Answer:
the dimensions that minimize the cost of the cylinder are R= 3.85 cm and L=12.88 cm
Step-by-step explanation:
since the volume of a cylinder is
V= π*R²*L → L =V/ (π*R²)
the cost function is
Cost = cost of side material * side area + cost of top and bottom material * top and bottom area
C = a* 2*π*R*L + b* 2*π*R²
replacing the value of L
C = a* 2*π*R* V/ (π*R²) + b* 2*π*R² = a* 2*V/R + b* 2*π*R²
then the optimal radius for minimum cost can be found when the derivative of the cost with respect to the radius equals 0 , then
dC/dR = -2*a*V/R² + 4*π*b*R = 0
4*π*b*R = 2*a*V/R²
R³ = a*V/(2*π*b)
R= ∛( a*V/(2*π*b))
replacing values
R= ∛( a*V/(2*π*b)) = ∛(0.03$/cm² * 600 cm³ /(2*π* 0.05$/cm²) )= 3.85 cm
then
L =V/ (π*R²) = 600 cm³/(π*(3.85 cm)²) = 12.88 cm
therefore the dimensions that minimize the cost of the cylinder are R= 3.85 cm and L=12.88 cm
it will take to fill a 10 liter container= 33.3 hours
