If a spherical ball is enlarged so that its surface area is 9 times greater than its original surface area, then the original radius was multiplied by _________.
Surface area is 4pir^3, so for the total to be enlarged by a factor of 9, the radius must go from r^3=1 to r^3=9, so the radius was multiplied by the cube root of 9.
\(4\pi r^2 \implies \) the original radius was multiplied by 3 times.
did you understand the explanations?
Sorry i just got back from eating o.o
good to have "food for thought" :)
yes and water xD
one says 9 the other says 3 o.o
ok, lets call the radius of the smaller sphere \(r_s\) and the radius of the bigger sphere \(r_b\). then their surface areas would be:\[S_s=4\pi r_s^2\]and:\[S_b=4\pi r_b^2\]agreed?
Answerthisnow: do you know the surface area formula?
ya
surface area for what?
lol
ok, so now if we take their ratios we get:\[\frac{S_b}{S_s}=\frac{4\pi r_b^2}{4\pi r_s^2}=\frac{r_b^2}{r_s^2}=(\frac{r_b}{r_s})^2\]agreed?
FFM: your brain is never foolish - its you who are foolish! :)
lolol alright so the answer is 9? my brain is half melted....
I was right. It is 3 indeed.
opps my brain broke for a second there sorry
Answerthisnow: do you agree with the ration formula I typed above?
*ratio
Yes you are right asnaseer, I should stop second guessing myself :P
yeah :)
i.e. we have shown that:\[\frac{S_b}{S_s}=(\frac{r_b}{r_s})^2\]
ok, so now we just take square roots of both sides to get:\[\sqrt{\frac{S_b}{S_s}}=\frac{r_b}{r_s}\]
and you are told that:\[\frac{S_b}{S_s}=9\]
so the answer is?
\[\frac{r_b}{r_s}=\sqrt{\frac{S_b}{S_s}}=\sqrt{9}=?\]
3
bingo!
Is there a short way of doing stuff like this? I have 50 question to go xDD well tecnically i have 110 to go
there is a general rule which says: 1. the ratio of areas is equal to the square of the ratio of lengths 2. the ratio of volumes is equal to the cube of the ratio of lengths but these can be derived as I did above.
Ok. Can i ask another here or should i just make another question thingy
Short way: Take the positive square root.
you should really post a new question "thingy" :)
kk xD i will thanks for the help
yw
Join our real-time social learning platform and learn together with your friends!