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]

Oct/Nov 2004. MCQ Q20.
q18

Q20

Pressure_p at depth 2x= atm+2xD_Pg

on the right arm  pressure         =atm+xD_Qg

they are at equilibrium
 atm+2xD_Pg = atm+ xD_Qg
simplifying on both side you get
2D_P= D_Q
^D_P/_DQ= 1/2

note D = density

[Hypernova]

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?

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$!$]

Change in KE is zero, since no chnage in velocity

using your equation

0+9000x40 = Work done by friction + 20000x12

its the same thing

Thanks

Q20

Pressure_p at depth 2x= atm+2xD_Pg

on the right arm  pressure         =atm+xD_Qg

they are at equilibrium
 atm+2xD_Pg = atm+ xD_Qg
simplifying on both side you get
2D_P= D_Q
^D_P/_DQ= 1/2

note D = density

Please explain this equation u used. We have not been taught it before..
Thank you

[ashish]

Thanks

Please explain this equation u used. We have not been taught it before..
Thank you

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 + 2xD_pg

pressure in the right arm (in the diagram the liquid has height x) = atm + xD_Qg

since pressure in a liquid i at same level(height) is equal

atm + 2xD_pg = atm + xD_Qg

D_p= density of liquid P
D_Q= density of liquid Q

hope it helps

[$!$RatJumper$!$]

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 + 2xD_pg

pressure in the right arm (in the diagram the liquid has height x) = atm + xD_Qg

since pressure in a liquid i at same level(height) is equal

atm + 2xD_pg = atm + xD_Qg

D_p= density of liquid P
D_Q= density of liquid Q

hope it helps

Thank you so much! +rep

can u please explain S10 P11 Q16

[Deadly_king]

can u please explain S10 P11 Q16

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$!$]

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
 

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]

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

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$!$]

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.

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?

[Deadly_king]

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?

Hmm...........this equation can be altered. You don't need to keep it fixed. It depends according to the question. I always develop my own equation in the exams though you must obey the principle of conservation of enrgy.

It's only used when you need the principle of conservation of energy.

[$!$RatJumper$!$]

So isnt energy being conserved in this system of the car?

[Deadly_king]

So isnt energy being conserved in this system of the car?

Not really since you're not told about frictional force.

[$!$RatJumper$!$]

i see ok thank you Can u help with W09 P11 Q 14, 22, 23, 26, 27, 28

Thank you.. and sorry if its a lot of questions

[Deadly_king]

i see ok thank you Can u help with W09 P11 Q 14, 22, 23, 26, 27, 28

Thank you.. and sorry if its a lot of questions

Nov 09 p11

11. The centre of gravity of an object is defined as the point at which the weight can be considered to act.

Answer is C

14. Initial k.E is given to be E.
Since the angle of projection is 45^o, the initial kinetic energy will divide equally to give the final P.E and the final K.E.

So final k.E will be 0.5E

Answer is A.

22. Youngs Modulus is given as Y = TL/Ae where
T : Tension in the string
L : Length of the string
A : Area of cross-section of the string.
e : extension produced.

You're asked to find the % length the string contracts. Hence you need to find the percentage of e/L which is also equal to F/AY. (from equation above)

Hence replace the values given to you to get 20/(pie*(diameter/2)²*2.0 x 10¹¹) x 100

It will come out to be 0.051%.

Hence answer is B

23. You should know this from the elctromagnetic spectrum.

Max wavelength of U.V = 10^-13 and c = 330ms^-1

Hence c = f(lambda) -----> f = 330/10^-13 = 10¹⁵ Hz

Answer is B

26. You should use the formula dsin(theta) = n(lambda) where d = 1/N. Since the beam of light is parallel, theta will be 90^o. You should also convert the wavelength into millimetres since N is given as 300 lines/mm.

(1/300) x sin 90 = n(450 x 10^-6) ----> n = 7.5
Hence there'll be 7.5 lambda above and 7.5 lambda below. This makes 15 lambda. Each lambda will form one maxima.

Answer is D

27. Force = Electric field x charge

Electric field = Force / charge = F/q

Answer is D

28. Equilibrium => Electric force = Weight

Hence QE = mg ----> Q/m = g/E

Electric force is upwards(opposite to the weight). Since it is attracted towards the positive plate, it has to be negatively charged.

Answer is B

Hope it helps