• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Physics: Post your doubts here!

H

hope4thebest

Messages
160
Reaction score
197
Points
38
TaffsAsLevel said:
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
Click to expand...

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
 
TaffsAsLevel

TaffsAsLevel

Messages
168
Reaction score
50
Points
38
hope4thebest said:
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
Click to expand...

Thank you buddy
 
A*(a*)

A*(a*)

Messages
434
Reaction score
237
Points
53
Antalya said:
Hey where did you get these questions from? They're pretty good
Click to expand...

Thats Cambridge Pre -U , another thing in CIE link, thats a level ahead of A level! I am doing it just for practice ,some of them are really harrrrrrrrrrd , but just for the sake of papers :p
 
Y

Yousif Mukkhtar

Messages
373
Reaction score
125
Points
53
Warrior66 said:
Q9: i found this answer posted by someone else, so not taking any credits! :D
"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! :D
Click to expand...

You helped me a lot. Thank you for the great help and may Allah shine you with all As this May june session
 
You must log in or register to reply here.
Top