A bucket holding water has a diameter of 10 in. The water in the bucket has a height of 4in. Charlie places 3 b***s in the bucket of water. Each ball has a diameter of 6 in. What is the height of the water after he puts the ball in the bucket? Round to the nearest whole number.
+118723 A bucket holding water has a diameter of 10 in. The water in the bucket has a height of 4in. Charlie places 3 b***s in the bucket of water. Each ball has a diameter of 6 in. What is the height of the water after he puts the ball in the bucket? Round to the nearest whole number.
The volume of 1 ball = (4/3)*pi r^3
So the volume of 3 b***s is 4pi r^3
Let V=4pi r^3 Where r=radius of ball You can work out what that is.
The surface area of the bucket is S = pi r^2 where r = radius of bucket, you can work that out
V=hS
change in height is H=V/S
You can now work out what the new height is. :)
+130511 Volume of the water in the bucket before the b***s are added =
pi * (d/2)^2 * h = pi *(10/2)^2 * 4 = pi * (5)^2 * 4 = 100pi in^3
Volume of the three b***s =
3 * (4/3) pi (d/2)^3 = 4pi ( 6/2)^3 = 4pi (3)^3 = 108 pi in^3
So...the total volume of the water and b***s = 208 pi in^3
And [ assuming the bucket is tall enough] ......the height of the water will be :
208pi = pi * (10/2)^2 *h
208 = 5^2 * h
208 = 25 *h
208/25 = h = about 8.32 in = about 8 in [rounded]
