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IM HAVING PROBLEM IN QS 8 as wellAsalaoalikum wr wb
anybody?
pls anybody helpppppppppppppppppppppppppppppppppppppppp
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IM HAVING PROBLEM IN QS 8 as wellAsalaoalikum wr wb
anybody?
IM HAVING PROBLEM IN QS 8 as well
pls anybody helpppppppppppppppppppppppppppppppppppppppp
Yes.ok so if an object is falling in water will upthrust still be lower than drag and weight would be the biggest force. the object is not floating.
ok thnx i have another question i just posted please loook at it.Yes.
If you increase the resistance of the variable resistor, the current decreases. When the current decreases, the p.d. across XY will decrease. Therefore, to balance the galvanometer you need more of the p.d. that is present in the lower circuit, so you shift the contact towards Y.http://www.xtremepapers.com/community/attachments/9702_w02_qp_1-pdf.11782/
PLEASEEEEEEEEEEEEE can any1 give explanation for q. 35!!!!!!
the ms says the ans is D
If you put height at the y-axis and time at the x-axis, then the velocity will be its gradient (height/time). Since the gradient is constant, this means the velocity is constant which is correct.http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_1.pdf
in question 12 why is A the asnwer. When an object falls with terminal speed its terminal speed should remain constant. so shouldnt the graph be a straight line showing the acceleration is also 0???
yeh but if the line will be fully staight what will the velocity be then. Will it be constant at that time too?If you put height at the y-axis and time at the x-axis, then the velocity will be its gradient (height/time). Since the gradient is constant, this means the velocity is constant which is correct.
If the line was fully straight the velocity would be zero because the gradient would be zero.yeh but if the line will be fully staight what will the velocity be then. Will it be constant at that time too?
i was so frustated aftr 8 dat i left the ppr and jumpd on another 1!and what about rest ? you know em all?
JazakAllah khairen..Waalikumasalaam!
View attachment 12290
not clear how 0.001 = 10ms :SNov 2010 P11 Q5
3000 rev/min
= 50 rev/sec
time period for one rev = 0.02s
For the wave to appear on a 10 cm screen, 0.02/10 = 0.002s.
Then you will divide by 2 for one pulse to appear = 0.001s or 10 ms/cm.
ANY1 HAVIN MS FOR SUMMER 2001?????????????????
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