SOLUTION: please show the working.
if 0.25 inches^3 of paint is applied to a circular disk (top only) with a diameter of 20 inches what is the thickness of the paint?
At noon car A is
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Question 419104: please show the working.
if 0.25 inches^3 of paint is applied to a circular disk (top only) with a diameter of 20 inches what is the thickness of the paint?
At noon car A is traveling north at 30 mph and is located 20 miles north of car B. Car B is traveling west at 50 mph. what is the distance between the cars at 12:45 pm, rounded to the nearest mile?
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
if 0.25 inches^3 of paint is applied to a circular disk (top only) with a diameter of 20 inches what is the thickness of the paint?
At noon car A is traveling north at 30 mph and is located 20 miles north of car B. Car B is traveling west at 50 mph. what is the distance between the cars at 12:45 pm, rounded to the nearest mile?
..
First problem:
Volume = Area*height or Area*thickness
thickness=volume/area
Area of a circle= pir^2=3.14*10^2=314 in^2
thickness=.25 in^3/314 in^2=.000796178 inches
ans:
Thickness of the paint=.000796178 inches
..
second problem:
Distance = speed*time of travel
At 12:45, car A will have traveled 30*(45/60)=90/4=45/2 miles which is added to the 20 miles it has already traveled for a total of 20+45/2 =42.5miles north.
At 12:45, car B will have traveled 50*(45/60)=75/2=37.5 miles west.
We now have a right triangle with legs of 42.5 and 37.5 miles.
Using the pathagorean theorem to solve:
x=sqrt(42.5^2+37.5^)=56.68 miles
ans:
Distance between the cars at 12:45 PM = 57 miles
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