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Physics: Post your doubts here!

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@baby cry (don't cry now) :D
ok here it is : u know how gain in p.e is equall to mgh
mg is 20k, height gained is 12
product is 240,000J

Kinetic Energy is constant hence no loss in K.E, so all the energy is lost as heat !
work done is F into d, f is 9k distance is 40m so the product is 360000J

subtract both u get 120000J
hence a is the answer
 
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yes i could figure that out.. but why do you subtract the potential energy.. can't you say the potential energy is the minimum work done??

thats the confusing bit..:p
 
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aaarghh why not? :mad:
what i got is that the work done on the body includes work done against friction as well. the potential energy gained at that instant needs to be subtracted from the total work done which leaves behind work done against friction heat...

and i didn't got ur 2nd statement. idk
 
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does any1 kno de conditions necessary to observe interference between two light sources?
Thank you :D
 
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Greetings guys! I'm confuse right now. how to differentiate whether a question ask us to use current r.m.s or peak current?

Thanks in advance.............
 
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dancrev992 said:
Greetings guys! I'm confuse right now. how to differentiate whether a question ask us to use current r.m.s or peak current?

Thanks in advance.............
Output of a typical power supply, ie 240-220 V is a rms value.
You will use the peak value if it states that 'taking maximum current' or something like that.
 
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babycry said:
yes i could figure that out.. but why do you subtract the potential energy.. can't you say the potential energy is the minimum work done??

thats the confusing bit..:p

The question will tell you that. :S Give an example of a question that confused you.
 
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its ok.. i understand now.. thanks for the reply.. and eid mubarak !!..

what you guys slaughtering?
 
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the electric potential at a point within a defined space is equal to the electric potential energy (measured in joules) at that location divided by the charge there (measured in coulombs). The electric potential at a specific location in the electric field is independent of qt. That is to say, it is a characteristic only of the electric field that is present. The electric potential can be calculated at a point in either a static (time-invariant) electric field or in a dynamic (varying with time) electric field at a specific time, and has the units of joules per coulomb, or volts.
source: Wikipedia
http://en.wikipedia.org/wiki/Electric_potential
 
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What is left hand rule used for and right hand rule used for? I am having trouble figuring it out. Also what is the right hand grip rule?
 

XPFMember

XPRS Moderator
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Assalamoalaikum!

Left Hand Rule is used when we supply the current! (force is produced)

Right Hand Rule is used when current is produced! (we provide force!)

Right Hand Grip Rule:
A quick way to work out the direction of the magnetic field in a solenoid is the right hand grip rule...

Make a fist and stick your thumb out (as if hitchhiking). Your fingers are wrapped in a circle, same as the coils in the solenoid. If you make your fingers point in the same direction as the conventional current around the coil - your thumb points towards the end of the solenoid that is the North pole.

image010.jpg
 
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hikarigenzo said:
hi,i really need your help. I really seem like giving up for physics. Can you explain to me about the question number 4,why when the time increases, the distance also increases? shouldn't it decreases?
Can you also explain to me c(ii)?
I'm really bad in physics, thanks
http://www.xtremepapers.com/CIE/Interna ... _qp_23.pdf

the time is increased, not the speed. so the distance increase with time
cii) s = (1/2) * a * (t^2)
s/(t^2)= (1/2) * a
s/(t^2) = gradient
twice of gradient = acceleration
 
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hikarigenzo said:
Hi,its me again. Due to different time zone, maybe you're all are sleeping. I will post all my doubt here. Hope some kind souls here will assist me.
http://www.xtremepapers.com/CIE/Interna ... _qp_23.pdf
it is the same question paper as before. Now is question 7a(ii),b(ii) and b(iii)
Thanks
7aii) 1) electric field strength = force/charge
E=F/q
F=Eq

2) for horizontal motion; s=vt
L=vt
L/v=t
bii) momentum change = force*time
= (Eq)*(L/v)
= (EqL)/v
 
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Hi,thanks for your help. But I still got question. From the question,d is the vertical distance between the base of the electromagnet and the bench. My question is, why d changes after the ball is released? The electromagnet and the bench is stationary. Seriously, I really do not understand about this question, assist is needed. Thanks again. Have a nice day.
 
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