Recent content by mbkau00

considering i am no more advanced in solving this problem than when I posted it; i think it is time to surrender and accept defeat!

am i right in saying this: if the ice block is fully submerged, the Volume displaced will equal the volume of the ice block. if the ice block is somewhat submerged, the volume displced will be less than the volume of the ice block

Sorry, I meant pressure will vary, not density. Ok if the ice block will displace it's own mass (100g) in a solution, it will displace 100ml if the densities were the same? As the density of the fluid is 1100kg m-3, whereas the iceblock is 900kg m-3, it won't be 100ml displaced. As named...

Regarding pressure - depth standpoint: i understand that density will increase as the object decreases vertically. no change horizontally at a fixed depth. thanks for your help.

Ok I am very new to physics (3 weeks) so i don't know or understand what calculations will keep this straight forward; as i still don't have a formula to work with. nonetheless i'll try and workout what you mean: - ice is .1kg - how much solution needs to be disaplced to keep forces...

I'm not really sure. I'll guess that it would almost be fully submerged; and as the ice block weighs less than the fluid it would float. By how much I don't know. Yes I didn't look at that calculation like that. It isn't ^3, only ^2.

Thank you.

Your analogy definitely adds a great perspective. Thanks so much. I still can't see how the answer is 70% of charge went from C1 to C2.

1.5*10^-7Q1 = 1*10^-15 - 5*10^-8Q1 Q1 = 6.67*10^-9 2*10^-15 - 1*10^-7Q2 = 5*10^-8Q2 Q2 = 1.33*10^-8 I am honestly as lost as when I started as I don't have a definitive approach in why certain variables are, or aren't related.

Thanks for your help and persistence. Unfortunately I don't get an answer near 70% which is the correct answer. I think i must admit defeat, considering time spent. Thanks again :)

rearranging: V=Q1/C1 V=Q2/C2 q1+q2= 2x10^(-8) rearranging and substituting: V=q1/0.05uF, 0.05uF*V=q1 V=q2/0.1uF, 0.1uF*V=q2 0.05uF*V+0.1uF*V = 2x10^-8

doesn't V need to be solved for? substituting q1 into the last equation and solving for q2 will equal 0. so to clarify: c1 = 0.05uF c2 = 0.1uF q1 = 2x10^-8 V is unknown q2 is unknown

The equations being? V=Q1/C1 V=Q2/C2 q1+q2= 2x10^(-8)

the charge on q1? 2x10^-8C q1+q2 = 2x10^(-8)C

sorry i don't know what the sum of charges equal. q1+q2 = ?