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No it was 4.8Nit is 5N for the tension
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No it was 4.8Nit is 5N for the tension
I think you are missing one thing from your equation mgsinO becaause it was a hill and its force had to be taken into account in the equation.HOW?\
WELL GUYS FOR THIS QUES WE WERE GIVEN ACCELERATIONS AT TOP AND BOTTOM RITE FOR EG LETS TAKE ONE SPEED THE ONE AT BOTTOM WE HAD TO CALCULATE ITFIRST
LOOK WE APPLIED F=MA RITE BY WHICH F-1000(RESISTIVE FORCE)=1250x4 F=6000 THAT IS THE DRIVING FORCE NOW FORMULA POWER=FORCEXVELOCITY 30000W/6000=5M/S THAT IS THE VELOCITY AT BOTTTOM THEN FOR THE SAME AT THE TOP BUT THIS TIME USE A AS 0.2 THEN SOLVE IT
and mgsinO was exactly 750NI think you are missing one thing from your equation mgsinO becaause it was a hill and its force had to be taken into account in the equation.
i love you manand mgsinO was exactly 750N
hmm thx for clearing this up. yup my fault srry guys but how many marks will i loose ? DAMN ! frgt thisFor B: T - 10(1-m) = a(1-m) <--------you forgot this
T - 10 +10m = 2 - 2m
T = 12 - 12m
Replace in A
8m = 12 - 12m
20m = 12
m = 0.6 kg
hehe thank youi love you man
well can u cnfrm me abt all my answers my Q6 is wrong wht abt the else ? hmm n for coefficient pasrt from rite we had angle 30 in part a hmm then we calculated the forward component of weight we had the acceleration so it was forward-friction=mxa then we got friction then divide by normal component if u have dont the same still answers diffeerent then post me q values ll do ithow did you get the value of coefficient of friction in Q5? mine was .08 something
and work done in Q6?
n my q 6 is wrong well i didnt add up the x component of weight so how many marks i wld loose? in later in part two ?how did you get the value of coefficient of friction in Q5? mine was .08 something
and work done in Q6?
not many,dont worry.you just missed out one valuen my q 6 is wrong well i didnt add up the x component of weight so how many marks i wld loose? in later in part two ?
i think dat coefficient of friction was "0.9"1. 1090J
2. a=2m/s2 m=0.6kg T=4.8N
3. 10m/s a=0.4m/s2
4. 39N direction 22.6' AC from x direction
5. 0=30' coefficient=0.693
6. gain in KE=344000J driving force workdone= 1220000J
7. distance OA=216m a=-0.72m/s2 u=17.6m/s
an get me all values i can tell the answer b calculationsi think dat coefficient of friction was "0.9"
i dont remember all the values.....if u c
an get me all values i can tell the answer b calculations
i dont remember all the values.....
u=4.4 m/s
v=0
s=65m
m was not given
t= (i dont remember, u?)
angle was 3o
man how did u solve that kinectic energy question...did u consider the weight component in frictional force or not....cuz last time it was the same qustion and i forgot to take that in account and got it right...this time i was confused...?
no you are absolutely wrong,you must have made a mistake I got 2 different values.And certainly weight component has to be taken into account!! then what is the use of including hill(inclined plain) in this question??i think that weight component is not taken into account while calculating energy....
during the paper i first calculated it with the weight component and the speed at bottom and at top were both 0.14 m/s
mass was only given to calculate energy
no you are absolutely wrong,you must have made a mistake I got 2 different values.And certainly weight component has to be taken into account!! then what is the use of including hill(inclined plain) in this question??
no brother you are mistaken
yep i did it this way but most people did mistake like they did the last time...i think that weight component is not taken into account while calculating energy....
during the paper i first calculated it with the weight component and the speed at bottom and at top were both 0.14 m/s
mass was only given to calculate energy
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