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Page history last edited by PBworks 14 years, 3 months ago

Problem 1 of ?


A spherical weather balloon has diameter 3.6 m. An additional 20m3 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.






The first thing that we have to do is find the original volume of the balloon by using the formula V=(4*pie*r3)/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*r3)/3.


V= (4*pie*r3)/3

v= (4*pie*(1.8)3)/3

V= 24.43 m3


The probem tells us that an additional 20m3 of air is pumped into the ballon, therefore we have to add 20m3 into our original volume.


24.43m3 + 20m3 = 44.43m3


The new volume of the balloon is 44.43m3




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


V= (4*pie*r3)/3

44.43m3 = (4*pie*r3)/3 Cross multiply

4*pie*r3 = 44.43m3*3

4*pie*r3 = 133.29m3 Divide both sides by 4pie to isolate r

r3= 10.61m3 Take the Cube roots of both sides

r= 2.19m


The new radius of the balloon is 2.19m




The formula for the surface area of a cylindef is S.A.= 4*pie*r2.


S.A.= 4*pie*r2

S.A.= 4*pie*(2.19)2

S.A.= 60.27m2


The new surface area of the ballon is now 60.27m2


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

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



Periodic Functions

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