Physics: Post your doubts here!

[talha-_-khan]

The piston is in equilibrium, as it is not moving. Thus outside pressure must equal the pressure inside. Thus force on piston
100000 * .003= 300N

Now, the area of the cylinder doubles, which halves the pressure inside it. Thus, the force on the piston now would be

50000* .003 = 150N

The reason i was not replying to your question was because i was confused about how the piston would remain in its place in diagram 2 as the air pressure outside is now more than the pressure inside. However, i had not read the question carefully enough as it said "held at". Sorry for the late reply

Thanks

 

[Suchal Riaz]

Snow Angel

 

[Hijab]

If i am right, the circuit is like this
View attachment 39203


The current divides at A. Thus the two resistors from A to B and B to C are in series with each other, and the resistor connected diagonally is parallel with them. So,

1/r = 1/6 + 1/12
r = 4 ohm

Now after this, The next resistor between C and D is in series once again. Thus the resistance till here in the lower two loops become 4 + 6 = 10. Add it with the resistor in the upper loop

1/10 + 1/6
r = 3.75

The second part is correct.... The first part is wrong... As the answer is 3 ohms... But thankyou for ur help...

 

[usama321]

The second part is correct.... The first part is wrong... As the answer is 3 ohms... But thankyou for ur help...

Well, i only answered the second part For the first part, it would
1/12 + 1/6 + 1/12 = 3

 

[usama321]

The second part is correct.... The first part is wrong... As the answer is 3 ohms... But thankyou for ur help...

For the first part, the circuit is:
the upper two resistors are in series
The lower two resistors are in series
And the the one in the middle is well.. alone xD
And all the above are in parallel to each other
I made that diagram just for the second part, it is not applicable to part i

 

[biscuitbiscuit]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w05_qp_2.pdf
"Q7 b(ii)1", why is it 2.7 when the volt metre says zero

 

[usama321]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w05_qp_2.pdf
"Q7 b(ii)1", why is it 2.7 when the volt metre says zero

The voltmeter measures the potential difference between two points. the potential at both the terminals of the volt meter is 2.7. There is a difference of 0V

 

[biscuitbiscuit]

The voltmeter measures the potential difference between two points. the potential at both the terminals of the volt meter is 2.7. There is a difference of 0V

Ahhh!!!! this was so easy, thank you anyways

 

[biscuitbiscuit]

The voltmeter measures the potential difference between two points. the potential at both the terminals of the volt meter is 2.7. There is a difference of 0V

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w09_qp_21.pdf
Can you explain Q6(d), please?

 

[usama321]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_21.pdf
Can you explain Q6(d), please?

Using the answer in b as they ask in the question, we can see that more power would be dessipated in the circuit, and less power lost in the internal resistance of the battery. Thus less energy would be lost

 

[biscuitbiscuit]

Using the answer in b as they ask in the question, we can see that more power would be dessipated in the circuit, and less power lost in the internal resistance of the battery. Thus less energy would be lost

Still didn't get it, the graph shows the power dissipation decreases! as the value of X increases

 

[usama321]

Still didn't get it, the graph shows the power dissipation decreases! as the value of X increases

It would decrease, but the overall power produced in the cell would also decrease. As a result, it would get a greater portion of the power produced in the cell.

 

[usama321]

Still didn't get it, the graph shows the power dissipation decreases! as the value of X increases

Nah, wait a second, i think i am wrong. Lemme see it again

 

[usama321]

Still didn't get it, the graph shows the power dissipation decreases! as the value of X increases

Ok. With increased resistance, less power would be produced in the cell. Now, due to less power being produced, there would be less current in the circuit. This would lead to now lesser power being dessipated in the internal resistor due to the fact that there is lesser current in the circuit, and thus less energy being wasted

 

[Suchal Riaz]

Snow Angel

 

[biscuitbiscuit]

Ok. With increased resistance, less power would be produced in the cell. Now, due to less power being produced, there would be less current in the circuit. This would lead to now lesser power being dessipated in the internal resistor due to the fact that there is lesser current in the circuit, and thus less energy being wasted

 

[Suchal Riaz]

Snow Angel

 

[Mimick]

Salam alaikum ...
I need a physics male teacher for AS level who lives in Jeddah..Only for chapter 'WAVES' and few questions in past papers...Need urgently..whoever knows pls give me his contact number...Thanks

 

[Mimick]

Salam alaikum everybody...
I need a physics male teacher for AS level who lives in Jeddah..Only for chapter 'WAVES' and few questions in past papers...Need urgently..whoever knows pls give me his contact number...Thanks

 

[Mimick]

Salam alaikum everybody...
I need a physics male teacher for AS level who lives in Jeddah..Only for chapter 'WAVES' and few questions in past papers...Need urgently..whoever knows pls give me his contact number...Thanks