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ALL CIE PHYSICS DOUBTS HERE !!!
$!$RatJumper$!$:
Thank you for that. It was really useful!
Though i have a doubt, the equation is:
Change in K.E + Useful work done = Total k.E in overcoming friction + Change in P.E
Why didnt you use the Change in K.E in this calculation?
ashish:
--- Quote from: thecandydoll on November 13, 2010, 07:02:25 am ---Oct/Nov 2004. MCQ Q20.
q18
--- End quote ---
Q20
Pressurep at depth 2x= atm+2xDPg
on the right arm pressure =atm+xDQg
they are at equilibrium
atm+2xDPg = atm+ xDQg
simplifying on both side you get
2DP= DQ
DP/DQ= 1/2
note D = density
Hypernova:
--- Quote from: $!$RatJumper$!$ on November 13, 2010, 08:48:14 am ---Thank you for that. It was really useful!
Though i have a doubt, the equation is:
Change in K.E + Useful work done = Total k.E in overcoming friction + Change in P.E
Why didnt you use the Change in K.E in this calculation?
--- End quote ---
Change in KE is zero, since no chnage in velocity
using your equation
0+9000x40 = Work done by friction + 20000x12
its the same thing
$!$RatJumper$!$:
--- Quote from: Hypernova on November 13, 2010, 08:54:09 am ---Change in KE is zero, since no chnage in velocity
using your equation
0+9000x40 = Work done by friction + 20000x12
its the same thing
--- End quote ---
Thanks :)
--- Quote from: ashish on November 13, 2010, 08:50:56 am ---Q20
Pressurep at depth 2x= atm+2xDPg
on the right arm pressure =atm+xDQg
they are at equilibrium
atm+2xDPg = atm+ xDQg
simplifying on both side you get
2DP= DQ
DP/DQ= 1/2
note D = density
--- End quote ---
Please explain this equation u used. We have not been taught it before.. :(
Thank you
ashish:
--- Quote from: $!$RatJumper$!$ on November 13, 2010, 08:56:55 am ---Thanks :)
Please explain this equation u used. We have not been taught it before.. :(
Thank you
--- End quote ---
pressure in a liquid = h(rho)g
row means density which i represented by D
h=height
pressure at the same level(height) in a liquid is equal
pressure in left arm at a depth of 2x = atmospheric pressure (atm) + pressure by liquid
= atm + 2xDpg
pressure in the right arm (in the diagram the liquid has height x) = atm + xDQg
since pressure in a liquid i at same level(height) is equal
atm + 2xDpg = atm + xDQg
Dp= density of liquid P
DQ= density of liquid Q
hope it helps
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