Physics: Post your doubts here!

[hina khokhar]

Can anyone post notes on kinematic

 

[iwoeriuyure]

I got a B in AS level physics.with comonpent grades P1-B,P-2 A and a U in Practical.can i still get an Overall A in A level physics?

 

[periyasamy]

Guys for this question .Am i doing the right way?
First do relation that is resistance is inversely proportional to area.Then find area for both using the diameter given .Then relate the resistance with area.Is this rite?

 

[minie23]

http://olevel.sourceforge.net/papers/9702/9702_s13_qp_12.pdf

Kindly help me for question 14, 16, 17 please

 

[periyasamy]

http://olevel.sourceforge.net/papers/9702/9702_s13_qp_12.pdf

Kindly help me for question 14, 16, 17 please

For question 14--------
(0.35*3*9.81)+(0.1*1.4*9.81=(0.15*6*9.81)+ans
ans=2.8Nm

For question 16----is answer b.
if yes then this is the way
80*28/100=112/5
112/5 =potential energy
112/5=mgh
112/5=(o.12)(9.81)h,,,,,,,,,,,,,,,,,only 120 g taken into account as only the arrow has a movement.

For question 17--------mgsinteta*speed=40% of total power mgsinteta=force,,,,,,force*velocity=power
(total m)(9.81)sin30)==40%
100%=ans
Hope it helps.Please tell if the answer u get by this way is differnt from mark scheme...

 

[minie23]

For question 14--------
(0.35*3*9.81)+(0.1*1.4*9.81=(0.15*6*9.81)+ans
ans=2.8Nm

For question 16----is answer b.
if yes then this is the way
80*28/100=112/5
112/5 =potential energy
112/5=mgh
112/5=(o.12)(9.81)h,,,,,,,,,,,,,,,,,only 120 g taken into account as only the arrow has a movement.

For question 17--------mgsinteta*speed=40% of total power mgsinteta=force,,,,,,force*velocity=power
(total m)(9.81)sin30)==40%
100%=ans
Hope it helps.Please tell if the answer u get by this way is differnt from mark scheme...

Thank you loads !

 

[periyasamy]

Hai,guys need help here.Why is this object said to be not in equilibrium?

 

[minie23]

Hai,guys need help here.Why is this object said to be not in equilibrium?

Perhaps the pulling force is much larger than the weight !

 

[mynameisjh]

10 A stationary nucleus has nucleon number A.
The nucleus decays by emitting a proton with speed v to form a new nucleus with speed u. The
new nucleus and the proton move away from one another in opposite directions.
Which equation gives v in terms of A and u?
A v = (A/4 - 1)u
B v = (A – 1)u
C v = Au
D v = (A + 1)u
The answer is B, can anybody please explain this to me ??!!

 

[snowbrood]

10 A stationary nucleus has nucleon number A.
The nucleus decays by emitting a proton with speed v to form a new nucleus with speed u. The
new nucleus and the proton move away from one another in opposite directions.
Which equation gives v in terms of A and u?
A v = (A/4 - 1)u
B v = (A – 1)u
C v = Au
D v = (A + 1)u
The answer is B, can anybody please explain this to me ??!!

i already solved these questions
some important points
the nucleus is initally at rest so its momentum is zero
mass of neutron is equal to proton(
mass of electrons is negligible
lets consider the mass of neutron or proton equal to n
A is neuclon number which is number of protons+neutrons
total number of protons+neutrons times the mass of one of them
total sum of momentum before decaying=after decaying

(nA)*0=n(A-1)u-(n*1)v .

signs change as they are moving in opposite direction
rearraging it gives n(A-1)u=nV
dividing each side with n gives
(A-1)u=V

 

[DumbIdiot]

10 A stationary nucleus has nucleon number A.
The nucleus decays by emitting a proton with speed v to form a new nucleus with speed u. The
new nucleus and the proton move away from one another in opposite directions.
Which equation gives v in terms of A and u?
A v = (A/4 - 1)u
B v = (A – 1)u
C v = Au
D v = (A + 1)u
The answer is B, can anybody please explain this to me ??!!

using POCM( principle.....con momentum)
initial p = final p
initial p of system = 0
final p must also be = o
mass of parent nucleus = Ay( where y is the unified atomic mass constant)
mass of proton = y
mass of daughter nucleus = (A-1)y
o = vy + (A-1)( y )(-u) ( -u cuz u and v are in opposite directions)
rearrange and u`ll end up with v=(A-1)u

u can ignore y as it`ll cancel out eventually

 

[minie23]

mn=(A−1)amump=1amu​

​

pi=pf

0=mnu−mpv

0=((A−1)amu )u−(1 amu) v

(1 amu) v=((A−1)amu)u

v=(A−1)u

 

[Jessie N.]

Help please QAQ

 

[Jessie N.]

Hai,guys need help here.Why is this object said to be not in equilibrium?

The forces don't pass through a same point so not in equilibrium.
Remember that forces passed through a same point are not necessarily equilibrium, but the forces in equilibrium must pass through a same point ^_^

 

[minie23]

http://olevel.sourceforge.net/papers/9702/9702_s13_qp_12.pdf

Someone please help me for no. 20
Jessie N.

 

[periyasamy]

Help please QAQ

I think this is the way.Not sure.
p=v^2/r
total r=1/r +1/r+1/r
r=r/3
as is parallel,,,the voltage is same======>p=v^2/r
12=v^2/(r/3)
v^2/r=p
v^2/r=12/3
=4
If the answer is rite,then this wrking should be rite,i guess.

 

[Jessie N.]

http://olevel.sourceforge.net/papers/9702/9702_s13_qp_12.pdf

Someone please help me for no. 20
Jessie N.

I've done it before >w<
https://www.xtremepapers.com/community/attachments/image-jpg.32952/

 

[periyasamy]

The forces don't pass through a same point so not in equilibrium.
Remember that forces passed through a same point are not necessarily equilibrium, but the forces in equilibrium must pass through a same point ^_^

Thanks a lot buddy,really understand..

 

[Jessie N.]

I think this is the way.Not sure.
p=v^2/r
total r=1/r +1/r+1/r
r=r/3
as is parallel,,,the voltage is same======>p=v^2/r
12=v^2/(r/3)
v^2/r=p
v^2/r=12/3
=4
If the answer is rite,then this wrking should be rite,i guess.

Answer is A :/

 

[periyasamy]

Ans

Answer is 2V :/

Oh,i am sorry.I will try again now.