ALL CIE PHYSICS DOUBTS HERE !!!

[Deadly_king]

On which formula is the one you stated based on?
thanx

Just the principle of conservation of energy.

[TJ-56]

Thank you all guys
and
WISH YOU ALL THE BEST OF LUCK FOR TOMORROW!

[$!$RatJumper$!$]

Is s = ut + (1/2 * a * t²) and v² = u² + (2 * a * s) only for vertical movement of an object?
And is speed = distance / time only for horizontal movement?

[Dania]

No, they can be used for either vertical or horizontal.
The only difference is that acceleration would be taken as -9.81 ms^-2 if the object is going up. If it moving horizontally, the acceleration can be calculated using v²=u² + 2as or any other motion equation regarding acceleration.

You should remember that speed is a scalar, and velocity is vector. In many questions, they will state whether it is speed or velocity. If they say it is speed, you should use the equation Speed=Distance/Time to find what it is you're looking for.

I hope I was helpful

[Deadly_king]

Is s = ut + (1/2 * a * t²) and v² = u² + (2 * a * s) only for vertical movement of an object?
And is speed = distance / time only for horizontal movement?

No, they can be used for either vertical or horizontal.
The only difference is that acceleration would be taken as -9.81 ms^-2 if the object is going up. If it moving horizontally, the acceleration can be calculated using v²=u² + 2as or any other motion equation regarding acceleration.

You should remember that speed is a scalar, and velocity is vector. In many questions, they will state whether it is speed or velocity. If they say it is speed, you should use the equation Speed=Distance/Time to find what it is you're looking for.

I hope I was helpful

Very good answer. 

I would just like to add that for any of the equations of motion to be valid, acceleration should be constant.

[$!$RatJumper$!$]

No, they can be used for either vertical or horizontal.
The only difference is that acceleration would be taken as -9.81 ms^-2 if the object is going up. If it moving horizontally, the acceleration can be calculated using v²=u² + 2as or any other motion equation regarding acceleration.

You should remember that speed is a scalar, and velocity is vector. In many questions, they will state whether it is speed or velocity. If they say it is speed, you should use the equation Speed=Distance/Time to find what it is you're looking for.

I hope I was helpful

Thank you! Best of luck for today

[Deadly_king]

Thank you! Best of luck for today

Good luck to you too buddy

Hope we are blessed with an easy paper

[$!$RatJumper$!$]

haha thanx man
Good luck to everyone here too! You all have been amazing in helping out. I've learnt so much here and cleared so many doubts

[ashish]

Refer to the attachment.
Please explain.

TO the Examiner sir a voltmeter with infinite resistance doesn't exist XD

it's pretty easy

3.6V is for the 2000ohm resistance
V= IR
3.6/2000=I
I=1.8*10^-3A

let resistance of the thermistor be T
R=((1/5000)+(1/T))^-1
R=5000T/T+5000

V=IR
2.4=1.8*10^-3 5000T/T+5000
when solving for T

you will get 1818.2 ohm

[Deadly_king]

TO the Examiner sir a voltmeter with infinite resistance doesn't exist XD

To ashish..........it sure does not exist in practice, but it does on theory.

But each time you make use of a voltmeter in a circuit, you should assume it has infinite resistance so that it draws no current.

[ashish]

To ashish..........it sure does not exist in practice, but it does on theory.

But each time you make use of a voltmeter in a circuit, you should assume it has infinite resistance so that it draws no current.

yeah but i even argued with a teacher saying him that i will believe him only if he shows me an ideal voltmeter
and he smiled at me xD

[Deadly_king]

yeah but i even argued with a teacher saying him that i will believe him only if he shows me an ideal voltmeter
and he smiled at me xD

Hahaha..................very smart way to baffle the teacher.

[thecandydoll]

Oct/Nov 2004. MCQ Q20.
q18

[Hypernova]

Oct/Nov 2004. MCQ Q20.
q18

q18

Your looking for the work done against the 9.0kN force.
Work done by a force = Force x distance moved in the direction of the force

The body has not gained speed, this means that the work done by the 9.0 kN is equal to the work opposing it.
If the body had gained speed, then the work done by the 9.0 kN is greater than the work done opposing by an amount equal to the gain in kinetic energy

the 9.0 kN is equal to the work opposing it.
so the work done by the 9kN is equal to the workdone by friction and by its weight.


Work done by 9kN = Work done by friction + delta Potential energy

9000x40 = Work done by friction + 20000x12

Work done by friction (heat dissipated) = 120000 J

Ans is A

[elemis]

+rep Hypernova