# Problem 1 of ?

A spherical weather balloon has diameter 3.6 m. An additional 20m^{3} of air is pumped into the balloon, increasing its volume and surface area.

(a) Determine the new volume of the balloon.

(b) Determine the new radius of the balloon.

(c) Determine the new surface area of the balloon.

**SOLUTIONS TO THE PROBLEM**

A)

The first thing that we have to do is find the original volume of the balloon by using the formula V=(4*pie*r^{3})/3, but we have to find radius first.

To find the radius we have to divide the diameter by 2.

r = (diameter)/2

r = 3.6/2

r = 1.8

Now that we have the radius we can now use the formula V=(4*pie*r^{3})/3.

V= (4*pie*r^{3})/3

v= (4*pie*(1.8)^{3})/3

V= 24.43 m^{3}

The probem tells us that an additional 20m^{3} of air is pumped into the ballon, therefore we have to add 20m^{3} into our original volume.

24.43m^{3} + 20m^{3} = 44.43m^{3}

**The new volume of the balloon is 44.43m**^{3}

B)

To solve this problem we have to write our formula and put in what we have.

V= (4*pie*r^{3})/3

44.43m^{3} = (4*pie*r^{3})/3 **Cross multiply**

4*pie*r^{3} = 44.43m^{3}*3

4*pie*r^{3} = 133.29m^{3} **Divide both sides by 4pie to isolate r**

r^{3}= 10.61m^{3} **Take the Cube roots of both sides**

r= 2.19m

**The new radius of the balloon is 2.19m**

C)

The formula for the surface area of a cylindef is S.A.= 4*pie*r^{2}.

S.A.= 4*pie*r^{2}

S.A.= 4*pie*(2.19)^{2}

S.A.= 60.27m^{2}

**The new surface area of the ballon is now 60.27m**^{2}

**This shows the solutions for question A**http://am40s.pbwiki.com/f/pb.wiki.doc

**This shows the solutions for question B and C**http://am40s.pbwiki.com/f/pb.wiki2.doc

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