Physics: Post your doubts here!

[Thought blocker]

can someone help me out wiz ziz question plz.

An early experimenter, working in other than Si units, obtained the following ten values for the magnitudes of the changes on small oil drops:
9.82, 19.64, 39.28, 39.28, 34.37, 19.64, 19.64, 29.46, 19.64, 39.28.
What value do these results suggest for the magnitude of the charge of the electron as measured in these units?
A 2.45
B 4.91
C 9.82
D 19.64

I dont get the question.
Moreover this is the first time I have came across this type. Which paper it is!

 

[Czechoslovakia]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_11.pdf
question 14 - pleaseee! Thought blocker

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_11.pdf
question 14 - pleaseee! Thought blocker

14)
sagar had explained v.well..
Suppose we take the upward direction to be positive, and the downward direction to be negative, we can again write the momentum equations and use them to find the final velocity of the system - the forces between the clay and the lead pellet are huge compared to other external forces, so even though there are external forces acting on the system during the collision, we can assume the momentum stays approximately constant.

So, the initial momentum is the momentum of the bullet alone, which is equal to mv = (5.0/1000) * 200 = 1.0 kg ms^-1 . The clay block is stationary, so it does not contribute any momentum to the system initially.
When the bullet collides with the clay, the bullet gets stuck in the clay and they both move off with the same velocity, which we'll call v(f). The mass of the lead pellet+ the clay block = (95/1000 + 5/1000) = (100/1000) = 0.1 kg.
Their final velocity = v(f)
Therefore, the final momentum of the system is 0.1 * v(f)

Since this is equal to the initial momentum, we can write (0.1 kg) * v(f) = (1.0 kg ms^-1) and so v(f) = 10 ms^-1.

This is the speed with which they rise after the collision. While the system moves upwards, gravity accelerates them downwards at a constant rate of -9.81 ms^-2. When they reach the highest point above the original position, their velocity is 0. Therefore, we can use the formula v^2 = u^2 + 2as to give us "s", which is the maximum height. So,

(0)^2 = (10)^2 +2(-9.81)s
19.62 * s = 100
s = 5.09 m = 5.1 meters = A.

 

[aliciaa]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w09_qp_41.pdf
No. 8 (c)(ii). Can someone explain how do i solve this please? I checked the ms but i didnt understand where did 28.28 come from?

 

[Thought blocker]

See (last) comment at http://physics-ref.blogspot.com/2014/05/physics-9702-notes-worked-solutions-for.html

This link is really helpful
Zepudee it will help you a lot. Keep visiting this blog.

 

[manutd96]

pls help me with this thx

 

[princess Anu]

Somebody please help me out with this!!

Q: The pressure p and Volume V for a given mass of gas may be given by (p+a/v2) (V-b) = c.
Find the units of a, b and c.

 

[Thought blocker]

pls help me with this thx View attachment 46948

Somebody please help me out with this!!

Q: The pressure p and Volume V for a given mass of gas may be given by (p+a/v2) (V-b) = c.
Find the units of a, b and c.

Zepudee

 

[Thought blocker]

Somebody please help me out with this!!

Q: The pressure p and Volume V for a given mass of gas may be given by (p+a/v2) (V-b) = c.
Find the units of a, b and c.

You can't add a pressure to a temperature, it does not make sense. So the a/v2 must have units of pressure if it is to be added to p. Similarly b must have units of volume.
Therefore, b has units= m^3
a/v^2 must have unit the same as pressure so,
N/m^2 = a/(m^3)^2
a =Nm^6/m2
a =Nm^4
c = N/m^2 x m^3 = Nm

 

[princess Anu]

You can't add a pressure to a temperature, it does not make sense. So the a/v2 must have units of pressure if it is to be added to p. Similarly b must have units of volume.
Therefore, b has units= m^3
a/v^2 must have unit the same as pressure so,
N/m^2 = a/(m^3)^2
a =Nm^6/m2
a =Nm^4
c = N/m^2 x m^3 = Nm

Thanks a lot.
May Allah bless you.

 

[Duaa Akhtar]

Hey can I get an A* in phy if I got A in P1 and P2 but B in practical? and how plzzzzz???

 

[Thought blocker]

Hey can I get an A* in phy if I got A in P1 and P2 but B in practical? and how plzzzzz???

Yes you can get A* with this As level grades.
What was that "how" for?

 

[ameeen]

AOA.
i am stuck in multiplication of vectors section can any one has the notes of cross and dot product so kiand send me
thanks

 

[Thought blocker]

AOA.
i am stuck in multiplication of vectors section can any one has the notes of cross and dot product so kiand send me
thanks

This is physics thread, I guess.

 

[akshay 999]

I dont get the question.
Moreover this is the first time I have came across this type. Which paper it is!

November 2000 paper 1 question number 27

 

[Thought blocker]

November 2000 paper 1 question number 27

https://www.xtremepapers.com/commun...st-your-doubts-here.9860/page-592#post-850697

 

[akshay 999]

Ans C. All other values are (integral) multiple of it. A and B are too small - no info about them from values given

THere was once are question like in multiple choice - one of the recent years. I don't remember which one though.

No its not C but B i have check it in the answer sheet.

 

[The Godfather]

just use the cambridge application booklet + see the worked solutions from the link in my signature below.

this should be enough

If this was enough, everyone would have touch the A*s!

 

[The Godfather]

if you don't get an A* after studying these notes and practicing all these past exam questions (you are even provided with their solutions there), then REALLY i don't know what to do to obtain an A*.

I need notes apart from textbook and past papers doesn't help to score A* if it does then all the candidates SHOULD have had score A*.

 

[Duaa Akhtar]

Yes you can get A* with this As level grades.
What was that "how" for?

i mean what things to emphasise more on to get an A*?