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Haha none are mind blowing I can do those, try my question june 2002 question 36. I dare ya
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Amyone to solve these challenging, mind-blowing question?
View attachment 28655
It is A, it should always be perpendicular to surface and arrow towards negative, but then the field is not uniform and closest means stronger field hence closer lines
Help me with nov 2002 question 36 please
It is A, it should always be perpendicular to surface and arrow towards negative, but then the field is not uniform and closest means stronger field hence closer lines
Help me with nov 2002 question 36 please
look as this is a parallel circuit...both terminals will have 2 V potential difference across them.Resistors have same value(5 ohm) so p.d will divide by 3 for each resistor that is 2/3 V.After passing 1st resistor in 1st series p.d remaining would be (2 - 2/3=4/3) at X. In 2nd series current passes through 2 resistors before coming to Y. P.d at Y would be (2 -2/3 -2/3= 2/3). So p.d between X and Y would be ( 4/3 - 2/3 = 2/3). A answer
E=mgh
Hey where did you get these questions from? They're pretty good
Hey where did you get these questions from? They're pretty good
Hey where did you get these questions from? They're pretty good
Hey where did you get these questions from? They're pretty good
Tougher than CIE?Check OCR they have way tougher questions, more mind twisting
Thank you! This is really helping me practice. I'm sure you'll nail an A*!http://papers.xtremepapers.com/CIE/Cambridge Pre-U/ physics section, only some of them are there :/
Q9: i found this answer posted by someone else, so not taking any credits!
"initial momentum of the ball is 2mv before collision... if collision was perfectly elastic one, all the momentum would have been gained by the ball... and this would have made the change in the momentum equal to 4mv.. 2mv-(-2mv)... now it is given that collision is inelastic, it will lose some of its momentum to the wall.. so now change in momentum will be less than 4mv but cant be less than the initial momentum... so the answer is C.. i think you cannot verify the answer by the calculations..."
Q11:
The tension in the rope needs to balance out the weight of the student acting downwards, so since the total tension needs to be opposite and equal to the weight, Let tension in rope on left side be T1 and the tension on rope on right side T2 and :. Total tension = T1sino + T2sino
add them up... W= 2Tsino, rearrange to get T= W/ 2sino. Answer: B
Q14:
Weight of box on slope with inclination of 30 degrees, mgsino= 200*sin30 and this acts downwards along the slope and the frictional force also act downwards along the slope so...(200*sin30) + 150= 250
And since W=F*d, we have to find out the distance d along the slope, so hypotenuse...1.5/sin30
The forward force to maintain a steady speed is the frictional force plus the component of weight along slope
W= ((200*sin30) + 150)*(1.5/sin30 )
Answer: D
Hope this helps!
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