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Physics: Post your doubts here!

daredevil

daredevil

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asd said:
Oh, you agree with me. :p ^_^ ^_- :ROFLMAO:
Click to expand...
do u get wat he is saying about making the graph of resistors? because i don't -__- as far as that graph is given in the question the resistance gradient is way too high now to extend the graph ithink... the graph is almost vertical even before reaching 6V ..... :O
 
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lubna1232

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daredevil said:
omigosh.. iwud never have been able to do it without the ms -___- :O

anyways u know that acceleration due togravity is 9.81ms^-2
and acceleration is in the downward direction so it will be 9.81sin15
then apply a formula in which u don't need to put in the final velocity because u don't know it and u can calculate the time simultaneously.
from the three eq of motion the formla we have is:
s=ut + (1/2)at^2

as it is initially at rest so u=0
s=at^2 is the eq u hav
substitute the values and take it from there :)
Click to expand...

thank yoouu so much :D part a iii of the same question is even weirder :/ can you try that one as well?
 
A

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daredevil said:
do u get wat he is saying about making the graph of resistors? because i don't -__- as far as that graph is given in the question the resistance gradient is way too high now to extend the graph ithink... the graph is almost vertical even before reaching 6V ..... :O
Click to expand...
asd said:
Dude, Work done by "asd" per unit time is the same as (ten to the power of infinity Newtons)*(speed of light)
Click to expand...
will u both tel me the questio u both sound like talking in french :p
 
asd

asd

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daredevil said:
do u get wat he is saying about making the graph of resistors? because i don't -__- as far as that graph is given in the question the resistance gradient is way too high now to extend the graph ithink... the graph is almost vertical even before reaching 6V ..... :O
Click to expand...
yeah, me neither.
well, his assumption is correct in a sense that Pd across C would be higher than that across R. but then again I fear the examiner isnt going to accept it cause the actual pd across C is going to be much more than that (as you said, the curve is almost vertical right after 6 V)
 
daredevil

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daredevil said:
np.... and ok lemme check and i'll get back atcha if i can do it :p
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yahan pe to i think u can use pretty much any formula in the book.... becuase u have the time, the acceleration, the distance moved, and the initial velocity and need to find the final velocity...
try it with 2as=v^2 - u^2

then do tell me if u get the answer... i cna't solve it ryt now because i am entangled in another hell of a question -______-
 
daredevil

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asd said:
yeah, me neither.
well, his assumption is correct in a sense that Pd across C would be higher than that across R. but then again I fear the examiner isnt going to accept it cause the actual pd across C is going to be much more than that (as you said, the curve is almost vertical right after 6 V)
Click to expand...
as far as the quality of the assumption goes he is absolutely correct like it should be more than 1500 but how can he tell that it is downright 2000 is past me.... uffff!!
 
asd

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daredevil said:
as far as the quality of the assumption goes he is absolutely correct like it should be more than 1500 but how can he tell that it is downright 2000 is past me.... uffff!!
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LOL jee jee aunty main samajh gia. :ROFLMAO:
he said he was just "assuming" :D
 
lubna1232

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daredevil said:
yahan pe to i think u can use pretty much any formula in the book.... becuase u have the time, the acceleration, the distance moved, and the initial velocity and need to find the final velocity...
try it with 2as=v^2 - u^2

then do tell me if u get the answer... i cna't solve it ryt now because i am entangled in another hell of a question -______-
Click to expand...

yes i got the answers correct :D v is 2.26 and t=0.89
 
daredevil

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syed1995 said:
WHAT ARE YOU GUYS GOING ON ABOUT... It all seems like Chinese and Japanese to me -,- May God have mercy on me!

A star , daredevil and asd
Click to expand...
hahaha...
yaar aik question hai u just hav to look through the thread to find out wat it was ... it is one hell of a troublesome question so plz plz plz find it and help syed !! its really confusing and i'm near pulling my hair out :O
 
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