Qualification > Sciences
ALL CIE PHYSICS DOUBTS HERE !!!
$!$RatJumper$!$:
--- Quote from: ashish on November 13, 2010, 09:11:04 am ---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
--- End quote ---
Thank you so much! :) +rep
can u please explain S10 P11 Q16
Deadly_king:
--- Quote from: $!$RatJumper$!$ on November 13, 2010, 09:50:08 am ---can u please explain S10 P11 Q16
--- End quote ---
Efficiency = Output / Input
Energy input : Force x distance : F
Energy output : P.E + K.E
P.E = mgh where h = x sin(alpha)
P.E : sin(alpha)
K.E = 0
Therefore efficiency = sin(alpha) / F
cancel out. and Efficiency is found to be sin(alpha) / F
Hence answer is D
$!$RatJumper$!$:
--- Quote from: Deadly_king on November 13, 2010, 10:16:14 am ---Efficiency = Output / Input
Energy input : Force x distance : F
Energy output : P.E + K.E
P.E = mgh where h = x sin(alpha)
P.E : sin(alpha)
K.E = 0
Therefore efficiency = sin(alpha) / F
cancel out. and Efficiency is found to be sin(alpha) / F
Hence answer is D
--- End quote ---
What always confuses me is the fact that:
1) How to know which values to take as the Energy input and Energy output
2) Why did you ADD K.E and P.E in this case? Isnt the equation: Change in K.E + Useful work done = Total k.E in overcoming friction + Change in P.E
I dont see how they will be added.
3) The definition of internal energy is sum of K.E and P.E but the equation stated above says otherwise.
Im SO confused :(
Deadly_king:
--- Quote from: $!$RatJumper$!$ on November 13, 2010, 10:37:48 am ---What always confuses me is the fact that:
1) How to know which values to take as the Energy input and Energy output
2) Why did you ADD K.E and P.E in this case? Isnt the equation: Change in K.E + Useful work done = Total k.E in overcoming friction + Change in P.E
I dont see how they will be added.
3) The definition of internal energy is sum of K.E and P.E but the equation stated above says otherwise.
Im SO confused :(
--- End quote ---
Energy input is the energy required to move the car from its initial position to its destination.
Energy output is the energy the car possesses once it is at its destination.
Change on K.E is zero since velocity is constant which implies that change in velocity is zero.
I added K.E and P.E since these are the two forms of energies that the car possesses once at the top.
3. Internal energy does not apply here. ;)
$!$RatJumper$!$:
--- Quote from: Deadly_king on November 13, 2010, 10:50:43 am ---Energy input is the energy required to move the car from its initial position to its destination.
Energy output is the energy the car possesses once it is at its destination.
Change on K.E is zero since velocity is constant which implies that change in velocity is zero.
I added K.E and P.E since these are the two forms of energies that the car possesses once at the top.
3. Internal energy does not apply here. ;)
--- End quote ---
I see :) that makes it a lot clearer now :) so when would we use the equation Change in K.E + Useful work done = Total k.E in overcoming friction + Change in P.E... like for what types of questions?
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