Physics: Post your doubts here!

[Laveeza]

The cross-section of an Olympic-size swimming pool filled with water. It is not
drawn to scale. The density of the water is 1000kg m–3... Can someone explain how the answer is 3.4 KPa and not 3.5kPa. Q20 winter 2015 p12?

Beacuse probably you are taking g as 10 instead of 9.81

 

[Thelastmoment]

http://papers.gceguide.com/A Levels/Physics (9702)/9702_w15_qp_13.pdf Q 37, 29,27,14

You're supposed to know 14 by thoery,
That a COUPLE consists of (i)Same magnitude forces acting in opposite directions (ii) Therefore producing rotational motion

 

[amina1300]

You're supposed to know 14 by thoery,
That a COUPLE consists of (i)Same magnitude forces acting in opposite directions (ii) Therefore producing rotational motion

opps i meant Q 15 I knew this one it quite obvious. Thanku fr helping me Can u try 9,37,15?

 

[Farjad Ilyas]

opps i meant Q 15 I knew this one it quite obvious. Thanku fr helping me Can u try 9,37,15?

For Q9, use the equation of motion s= ut + 1/2 at^2
We are only considering vertical velocity at the moment so: s=8, u=0, a=9.81 and you have to find t. If you put these in the formula you'll get the time as 1.27s, so C is the answer.

Fpr Q15, the 4N force has no effect on the moment about the pivot since its acting in line with it.
Total anticlockwise moment= (3N x 2m) + (2sin(30) x 4 ) = 10
Total clockwise moment = 5m x F = 10
Thus, F=2N so the answer is B

Q37 is kida hard to explain, you basically have to use V=IR ultiple times. For the last two resistors the voltage across each is 1V. So the voltage across the resistor connected in parallel is also 2V. The current through this resistor must be 2A (I= V/R). Thus the current across the 2nd last horizontal resistor adds up to 3A and so the p.d across it is 3V. The last two resistors we talked about are connected in parallel to the first vertical resistor and so the p.d across it is 5V and the current flowing through it is 5A. This mean that the current flowing through he first resistor is 8A and the p.d across it is 8V. Excluuding the very first resistor, the circuit is a big parallel setup with a voltage of 5V across it. So the e.m.f must be 8V +5V = 13V so the ans is D

 

[learner_ar]

Hi there,
I need help with this mcq.
Correct ans is D

Attached is the image of question 36 from 9702_W16_11

Link: https://drive.google.com/file/d/0B4LUJjAvVTcWMWVuRnRJMHBGckk/view?usp=drivesdk

 

[Munib Ahmed]

does anyone know where to get the physics may/june 2017 paper 41. i know it's not yet uploaded but i was hoping if anyone of u knew how to get the paper

 

[Thelastmoment]

opps i meant Q 15 I knew this one it quite obvious. Thanku fr helping me Can u try 9,37,15?

Q37- Theory -Kirchoff's rule One and Two
Hope this helps
The rest I think Farjad is correct

 

[Thelastmoment]

does anyone know where to get the physics may/june 2017 paper 41. i know it's not yet uploaded but i was hoping if anyone of u knew how to get the paper

I don't know if that's possible.. although try this site justpastpapers.com it usually releases the papers 24h after the exam is conducted.

Lemme see if I can help.

 

[SohaibButt]

It's D need explanation

 

[Thelastmoment]

It's D need explanation

When a beta particle is emitted from the nucleus the nucleus has one more proton and one less neutron.And we know that an alpha makes it even lower..
So, A & B no way then to make sure it's the minimum reduction we choose D.
Because, 2 betas would remove more than one beta which btw is obvious ...

 

[SohaibButt]

When a beta particle is emitted from the nucleus the nucleus has one more proton and one less neutron.And we know that an alpha makes it even lower..
So, A & B no way then to make sure it's the minimum reduction we choose D.
Because, 2 betas would remove more than one beta which btw is obvious ...

Oh damn y didn't i focused thanks a lot!

 

[Zenn99]

Can someone pls tell me how to get the ans for Q13 in FEB/MARCH 2017 PHY MCQ paper?

 

[Zenn99]

the paper is here, can someone pls tell me how to get the ans for Q13 AND Q4

 

[amina1300]

the paper is here, can someone pls tell me how to get the ans for Q13 AND Q4

Q4:T=1/f 1/50000 and divide this f by 2. This makes 1 complete wavelength in 2 divisions and hence you get C as the answer.

 

[Aishayasin]

Question 6 Answeris B

 

[amina1300]

View attachment 62292 Question 6 Answeris B

Vertically:
s=ut + 0.5at^2
2=0+10*t^2
t= 0.32
horizontally:
s=ut+0.5at^2
a is 0
6/0.632 = v Answer is b

 

[Aishayasin]

Vertically:
s=ut + 0.5at^2
2=0+10*t^2
t= 0.32
horizontally:
s=ut+0.5at^2
a is 0
6/0.632 = v Answer is b

Can u explain it step by step .Thanks

 

[Aishayasin]

ANSWER IS B

 

[amina1300]

Can u explain it step by step .Thanks

Ok so the cyclist is moving in two dimensions - both horizontally and vertically. Since these two components of motion are independent of each other, two distinctly separate sets of equations are needed - one for the cyclist's horizontal motion and one for its vertical motion.

We take Vertical Motion 1st because we know that for a free-falling object acceleration is g and hence we calculate the time taken for it to fall.
This time is the same for the horizontal component to reach its end point so while for the Horizontal motion we take the distance mentioned as 6m and divide it by the time(a=0) and we get the answer. (s=ut) Note: I took g as 10ms-1

 

[amina1300]

View attachment 62295

ANSWER IS B

Horizontally:
The horizontal velocity remains same 4ms-1
Vertically:
v=u+at Note : acceleration is negative. as object is falling.
v=8+((-1.62)*9)= -6.58

Resultant final velocity:
sqrt(ans^2+4^2 ) = 7.7 ms^-1


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