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First get the diagonal distance of hill by using sin 30 = 1.5/h
Diagram in question is not to scale.
thank youQ8
answer is A because speed increases uniformly so constant acceleration
Q6 + 5
when adding or subtracting 2 quantities add absolute error
when mulitplying or dividing add percentage error
when a quantity is to the power of something multiply power to percentage error
welcomethank you
use the equation
WD= PE + KE + WD[by friction]
PE=mgh (the weight of the box is 200N so no need to find mg)
so 200*1.5=300J
KE= question says steady speed so no change in KE so KE is 0
WD by friction: 150 * 1.5/sin30 (to find horzontal distance moved bybox) =450J
so 300+450=750J
jazakaALLAH loadzzzz...going 4 da ppr best of luck 2 any1 whos giving their pprs plzz pray 4 meFirst get the diagonal distance of hill by using sin 30 = 1.5/h
h = 1.5/sin 30 = 3m
So the work done by driving force is fsin theta *d = 200 sin 30 * 3 = 300 N
Work done against frictional force = 150 * 3 = 450 N
So 300 + 450 = 750 N
ur nt done wid the yet??? 0.osory me dont know
jazakaALLAH loadzzzz...going 4 da ppr best of luck 2 any1 whos giving their pprs plzz pray 4 me
i didn't scale it i redrew it -_-
Walikum Salaam. Sorry. I couldn't solve it either. Nibz is not online, let me tag hamidali391. URGENT.
You can't assume angles in such cases to be 45 degree. Yes the lines are perpendicular to each other, it should be 45 degree. But use the given data to find it.
Walikum Salaam. Sorry. I couldn't solve it either. Nibz is not online, let me tag hamidali391. URGENT.
sry i wasnt dere...Can someone pls help with this?View attachment 13023
ans is C
Capricedcapri abdullah ash bloody_mary
lol i just came back it was gud AHur nt donw wid the pp
ur nt done wid the yet??? 0.o
no psry i wasnt dere...
oh great!lol i just came back it was gud AH
mine went great!! wbu??How was the exam people ?
mine went great!! wbu??
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