Question #83e60

1 Answer
Sep 6, 2017

#"a)"# Eight tennis balls

#"b)"# #"1."# #268.08# #"cm"^(3)#

#" " "2."# #3267.26# #"cm"^(3)#

#"c)"# #34.4%#

#"d)"# #1734.16# #"cm"^(2)#

Explanation:

#"a)"# The radius of both the cylindrical container and each tennis ball is #4# #"cm"#.

So, one ball can be filled at a time.

Also, as the radius is #4# #"cm"#, the diameter of each ball must be #8# #"cm"#.

If we had eight balls, they would occupy #8 times 8# #"cm"# #= 64# #"cm"# in length.

The length of the container is #65# #"cm"#, so we can exactly fit in it a maximum of eight tennis balls.

#"#

#"b)"#

#"1."# The volume of a sphere is given as #V = frac(4)(3) pi r^(3)#:

#Rightarrow "Volume of each tennis ball" = frac(4)(3) times pi times (4)^(3)# #"cm"^(3)#

#Rightarrow "Volume of each tennis ball" = frac(4)(3) times pi times 64# #"cm"^(3)#

#Rightarrow "Volume of each tennis ball" = frac(256)(3) times pi# #"cm"^(3)#

#Rightarrow "Volume of each tennis ball" = 268.08257311# #"cm"^(3)#

#therefore "Volume of each tennis ball" = 268.08# #"cm"^(3)#

#"#

#"2."# The volume of a cylinder is given as #V = pi r^(2) h#; where #h# is the height, or length, of the cylinder:

#Rightarrow "Volume of the cylindrical container" = pi times (4)^(2) times 65# #"cm"^(3)#

#Rightarrow "Volume of the cylindrical container" = pi times 16 times 65# #"cm"^(3)#

#Rightarrow "Volume of the cylindrical container" = pi times 1040# #"cm"^(3)#

#Rightarrow "Volume of the cylindrical container" = 3267.2563597# #"cm"^(3)#

#therefore "Volume of the cylindrical container" = 3267.26# #"cm"^(3)#

#"#

#"c)"# From part #"b."# #"1."#, the volume of each ball is around #268.08# #"cm"^(3)#.

So the volume of all eight balls that can fit in the container is #8 times 268.08# #"cm"^(3) = 2144.64# #"cm"^(3)#.

#frac(2144.64 " cm"^(3))(3267.26 " cm"^(3)) = 0.656#, or roughly #65.6%# of the container is occupied by the eight balls.

Therefore, #100% - 65.6% = 34.4%# of the container is not occupied by the balls.

#"#

#"d)"# The surface area of a cylinder is given as #A = 2 pi r h + 2 pi r^(2)#:

#Rightarrow "Surface area of the container" = 2 times pi times 4 times 65 + 2 times pi times 4^(2)# #"cm"^(2)#

#Rightarrow "Surface area of the container" = 520 times pi + 32 times pi# #"cm"^(2)#

#Rightarrow "Surface area of the container" = 552 times pi# #"cm"^(2)#

#Rightarrow "Surface area of the container" = 1734.1591448# #"cm"^(2)#

#therefore "Surface area of the container" = 1734.16# #"cm"^(2)#