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Physics MCQs thread.
- Thread starter Zishi
- Start date
well i at last found the answer....its A....BT i still dont undrstnd one thing....i supposed speed of...PROTON as X and He as Y...in that 2nd eq. realtive speed of approach thing,,,,when i subtracted x from y i gt the ryt anser bt when i did the other way around i got wrong....so i dnt knw why is it and wats the proper method
yea hassam i got it
see here
let pr. has speed x and he has y
then we have 1*0.1-4*0.05=1*x+4*y is 1st equation
as collision is elastic so ke is conserve then
0.5*1*0.1^2+0.5*4*0.05^2=0.5*1*x^2+0.5*4*y^2 2nd equation
solve two equation u will get the ans
am i write??
can u provide me other question????? except cie 2002-2010
if yes mail me at
[email protected] you will be thanks
see here
let pr. has speed x and he has y
then we have 1*0.1-4*0.05=1*x+4*y is 1st equation
as collision is elastic so ke is conserve then
0.5*1*0.1^2+0.5*4*0.05^2=0.5*1*x^2+0.5*4*y^2 2nd equation
solve two equation u will get the ans
am i write??
can u provide me other question????? except cie 2002-2010
if yes mail me at
[email protected] you will be thanks
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hassam said:
A is the answer.
Speed of approach = Speed of separation
Speed of approach = 0.100c - (-0.05c) = 0.15c
Speed of separation = 0.01c - (-0.140c) = 0.15c
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hassam said:
Checked it, but I was wondering if you could email me pdf files from wenku baidu...
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WHY ITS NOT D ???????but C
well for this question...i used this method ...is this correct.....to travel at constant speed of 20m/s means that 200N D.F is balanced with air drag....
when climbing on slope both air drag and component of weight down the slope will be acting....we add them up nad multiply by 20...ans comes 23.6kW...APPROX 24 kW
when climbing on slope both air drag and component of weight down the slope will be acting....we add them up nad multiply by 20...ans comes 23.6kW...APPROX 24 kW