Physics: Post your doubts here!

[Marinaa1998]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s09_qp_4.pdf
I need someone's help with Q10 (b) please!

 

[Maayee]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_4.pdf
I need someone's help with Q10 (b) please!

this is a non inverting amplifier so the formula for the gain is G=1+R1/R2
so 25=1+120/R
24R=120
R=5

 

[Marinaa1998]

this is a non inverting amplifier so the formula for the gain is G=1+R1/R2
so 25=1+120/R
24R=120
R=5

Thanks a lot for your help! Can you please help me with (c)(i) as well?

 

[Maayee]

Thanks a lot for your help! Can you please help me with (c)(i) as well?

sure
ok so u need to find Vout so the formula is Gain=Vout/Vin
u already know ur gain 25 and Vin as -0.08 so Vout is 25x0.08=2V

 

[Marinaa1998]

sure
ok so u need to find Vout so the formula is Gain=Vout/Vin
u already know ur gain 25 and Vin as -0.08 so Vout is 25x0.08=2V

thank u so much

 

[Iris Dylan Lane]

What happens to the spring constant if the cross-sectional area is doubled?

 

[Muhammad Amer]

Can anyone help me with this?

"A power supply of e.m.f. E and internal resistance R is connected to a 10 kΩ resistor. The current drawn from the power supply is 0.91 mA. A voltmeter of internal resistance 10 kΩ is connected in parallel with the 10 kΩ resistor, and reads 8.3 V. Calculate the e.m.f. and the internal resistance of the power supply."

This question should be answered using Kirchhoff's laws.

 

[The Sarcastic Retard]

What happens to the spring constant if the cross-sectional area is doubled?

Doubled.

 

[muhammadali233]

Physicist http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-158.html
Q-784 is done wrong
L = 1/2*gT² , if T--> T/2 => 1/2*g*(T/2)² = 1/2*g*T²/4 = L*1/4

Answer: 0.25 L (B)

 

[Maayee]

can someone pls help me with this..
http://papers.gceguide.com/A Levels/Physics (9702)/9702_w15_qp_43.pdf
Q10 a an b and also 12 c

 

[princess Anu]

Book says :

1.Back Emf seems to imply opposition, something not wanted, yet in a motor it is the induced emf that is responsible for the mechanical work output of the motor.
2.The workdone against the induced emf in a motor is the useful work output of the motor. Workdone against the resistance of the wires is wasted as thermal energy

Can someone explain please?
How is back emf responsible for the ouput of a motor? :S

 

[AnonymousX9]

Guys pls help, M/J 2009 P22 Questions 3(b), 4(b) and 7(a).. ??? Please

 

[Eugene99]

Which book should I chose for Physics?

 

[Muhammad Amer]

Which book should I chose for Physics?

Probably Cambridge International AS and A Level Physics Coursebook with CD-ROM

 

[Eugene99]

Probably Cambridge International AS and A Level Physics Coursebook with CD-ROM

which one is the recent edition? any idea?

 

[My Name]

which one is the recent edition? any idea?

That one is the recent one.Second edition.

 

[Marinaa1998]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s09_qp_4.pdf
help please! 6(b)(i) why does the fact that the force on the magnet is downwards mean that the force on the wire is upwards?

 

[The Sarcastic Retard]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_4.pdf
help please! 6(b)(i) why does the fact that the force on the magnet is downwards mean that the force on the wire is upwards?

Force on wire is upwards according to newtons third law.
Use Flemings left hand rule. Thumb will point upward, second finger in direction from x to y. So first finger will be from pole p to other pole. Thus pole p is north pole as magnetic field direction is from north to south.

 

[Marinaa1998]

Force on wire is upwards according to newtons third law.
Use Flemings left hand rule. Thumb will point upward, second finger in direction from x to y. So first finger will be from pole p to other pole. Thus pole p is north pole as magnetic field direction is from north to south.

Thanks a lot!!

 

[The Sarcastic Retard]

Thanks a lot!!

Its okay.