- Messages
- 676
- Reaction score
- 756
- Points
- 103
plz telll me the answers of q4 in which S/2 and T/2 was given
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)
s was 13mplz telll me the answers of q4 in which S/2 and T/2 was given
i dont remember but i thinh it was 2.46what was acceleration in question 1?
3.46what was acceleration in question 1?
I approvehere are the steps how to do it!! if any doubts ask me! View attachment 26044View attachment 26045View attachment 26044View attachment 26045
Power was 20KWin question 7 answer was 0.47 m
in q 4 last part speed was 2.65 and in first paprt distanc was 13 .tension in pb was 13..P was 12KW and steady speed was 13.3
w work done against resistance was 750 and the question we have to show kinetic energy
plz tell me other questions,i forgot them
No, after B touched the floor A started another journey with the V of B then stopped moving (v=0), you had to calculate that "journey" distance and add the distance given in the question (which was B's heigh above the floor)d
dont you think the initial velocity was zero and final velocity was 8 m/s
No, after B touched the floor A started another journey with the V of B then stopped moving (v=0), you had to calculate that "journey" distance and add the distance given in the question (which was B's heigh above the floor)
I approve
P.S: Instead of integrating Acceleration you could've gotten the equation of the line 20 < t < 26 using y=mx+c
don't know which is faster tho
yes it was 20KW not 20000KW, one of my friend rote 20000KW and went like this was the answer...Power was 20KW
but i can assure u that this required less thinking than y=mx+cI approve
P.S: Instead of integrating Acceleration you could've gotten the equation of the line 20 < t < 26 using y=mx+c
don't know which is faster tho
i got.. 1.58 m and was pretty sure that's correct.. as i finfished 15 minutes earlier i sat and recalculated the whole thing... cant remember the full, but the last part i found that:
due to B's motion, A was pulled from rest to the same speed of B, v^2=0^2+2as, so distance = 's'..
Then A travelled a bit more due to subsequent motion( or continuing motion but against gravity until it stopped): 0^2=u^2-2gs (here u of A = final speed of B just before hitting ground,) hence a futher 's' travelled.
total dist= s+s, which came 0.92m by A towards P for me.
A travelled 0.92 m towards P.
and dist from P (2.5-0.92) = 1.58 m... plz share wat others did..
*****And wat was the MARKS for this part????
For almost 10 years, the site XtremePapers has been trying very hard to serve its users.
However, we are now struggling to cover its operational costs due to unforeseen circumstances. If we helped you in any way, kindly contribute and be the part of this effort. No act of kindness, no matter how small, is ever wasted.
Click here to Donate Now