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for first part integrate the equation
u will get
s=2t^2 - t^4/64 now u are given with s put it in the equation or quadratic k master ho aap
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for first part integrate the equation
Make the graph first. Then it will be easy. Meaning do b first and then a
tyfor first part integrate the equation
u will get
s=2t^2 - t^4/64 now u are given with s put it in the equation or quadratic k master ho aap
Area under graph = Distanceso do u mean that we should caculate the total distance from graph :/?
Area under graph = Distance
No, when you make the graph, you'll get the idea of max velocity=15. Then 15/y=0.0375 <--- last portion of graph, where y : 400. So, you add 400+ 600+ 2600 to get total time. (x) in the graph ive drawn.so do u mean that we should caculate the total distance from graph :/?
I got it, ty.No, when you make the graph, you'll get the idea of max velocity=15. Then 15/y=0.0375 <--- last portion of graph, where y : 400. So, you add 400+ 600+ 2600 to get total time. (x) in the graph ive drawn.
No, when you make the graph, you'll get the idea of max velocity=15. Then 15/y=0.0375 <--- last portion of graph, where y : 400. So, you add 400+ 600+ 2600 to get total time. (x) in the graph ive drawn.
Q6 karde bhaiNo, when you make the graph, you'll get the idea of max velocity=15. Then 15/y=0.0375 <--- last portion of graph, where y : 400. So, you add 400+ 600+ 2600 to get total time. (x) in the graph ive drawn.
No, when you make the graph, you'll get the idea of max velocity=15. Then 15/y=0.0375 <--- last portion of graph, where y : 400. So, you add 400+ 600+ 2600 to get total time. (x) in the graph ive drawn.
for part 1Q6 bhi kardo
sketching os required in part two,so cant apply this on part one....we have to use the formula.Q6 karde bhai
Why isn't it work done by pulling force-work dong against resistance=change in mechanical energy?ok look for potential energy we'll use mgh
1.2*10*1.5
we calculated the tension 15 N
W=F*s
=15*1.5
=22.5
work donne by pulling force = gain in P.E -loss in K.E.
sketching os required in part two,so cant apply this on part one....we have to use the formula.
ryt.i think you are talking about 5 ._.
yup I want 6.i think you are talking about 5 ._.
actually in the question it was written fixed smoot pulley so i assume no friction will be thereWhy isn't it work done by pulling force-work dong against resistance=change in mechanical energy?
I got 5ryt.
Oh sorry what question? I kinda skipped over most of these comments cause I didn't have the timeactually in the question it was written fixed smoot pulley so i assume no friction will be there
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