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Question 12, 14, 26, 28, http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w09_qp_11.pdf
Thanks !!
Thanks !!
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Question 12, 14, 26, 28, http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_11.pdf
Thanks !!
12. You have to consider the angles between T1 and R1 and T2 and R2. T1 cos(angle) gives you R1. What happens to cos(angle) if the angle decreases?
14. At the highest point of a parabolic trajectory, the vertical component of the velocity = o. So the K.E. can come only from the horizontal component, which is constant throughout.
26. Use sin90=nxlambda, then multiply by 2 (because you're taking the maxima on both sides of the middle maximum) and add 1 (the middle maximum).
28. Weight acts downwards so since it's in equilibrium the electric force has to act upwards. Since the plate up is positive and attracting the drop, that drop has to have a negative charge.
I put in numbers for this. Say V = 6 V and the resistance of each resistor = 2 ohms. The PD across the combination QR = 1/3 x 6 = 2 V. The Power through resistor R = (2^2)/2=2.
14. So why is the answer 0.5E instead of E
28. Why g/E instead of E/g
Can you show me the morking for 14. I still dont get it.14. (cos45)^2.
28. Mass=F/a. F=Eq, so mass=Eq/g. Charge=q. Charge/mass=q/Eq/g=qg/Eq. q cancels out, g/E is left.
Can you please help me with number 15 http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_11.pdfI put in numbers for this. Say V = 6 V and the resistance of each resistor = 2 ohms. The PD across the combination QR = 1/3 x 6 = 2 V. The Power through resistor R = (2^2)/2=2.
Can you show me the morking for 14. I still dont get it.
Can you please help me with number 15 http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_11.pdf
use total momentum before collision=total momentum after collision
therefore, 2*2=1*v
therefore speed of 1 kg trolley is 4..
now u have mass and velocity of both trolleys , right ?
find kinetic energy of both ... and simply add them !
hope that helps !
BTW bro thankss alott.. this was really very helpful!
13. All you gotta do is choose a pivot and work out the moments on either side of it...
14. mgh=1/2m(v^2). mgh/2 = 1/2m(x^2). Eliminate m, g, and 1/2. Cross multiply.
15. Find the angle to the horizontal of the slope. Then find the component of the weight parallel to the slope. Then use F=ma to find the acceleration. Then use a suvat. You can also do it using 1/2m(v^2)=mgh.
34. Resistance = ql/A. A=lxl, so Resistance=ql/lxl. l and l cancel out; R=q/l. l^3=V, so l = cube root of V = V^(1/3), so R=q/V^(1/3).
For Q13. there is already a pivot, so why would we choose one? It doesn't work with the given one.
For Q.14 I have no idea what you did..
I understood Q15. thank you.
And also I do not know what you did in Q34.
Can you explain more please?
but we don't have the velocity for the crane :sP=Fv.
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