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Posted by Mohit Kumar 5 years ago (10496025)
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Posted by Rohan Sinha 4 years, 7 months ago (10435839)
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Sia ? 4 years, 7 months ago (6945213)
m = 70 kg
v = 9 kmph = 9 * 5/18 = 2.5 m/s
u = 18 kmph = 18 * 5/18 = 5 m/s
Work done = change in kinetic energy
= initial KE - final KE
= 1/2 m u² - 1/2 m v²
= 1/2 * 70 * 5² - 1/2 * 70 * 2.5²
= 875 J - 218.75 J = 656.25 J
The speed has come down to half its value. But the Kinetic energy becomes 218.75/875 = 1/4 th of inital KE.
Posted by Devanshi Bavaskar 5 years ago (10170724)
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Posted by Varun Lalwani 5 years ago (10489732)
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Posted by Puja P 5 years ago (9918032)
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Posted by Vedha ..... 4 years, 7 months ago (9408112)
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Sia ? 4 years, 7 months ago (6945213)
using diffrentiation method
x = 18t + 5t^2
dx / dt = 18 + 10t
v = 18 + 10t
t = 2 sec
v = 18 + 10 * 2
= 18 + 20
= 38 m / sec
t = 3 sec
v = 18 + 10 * 3
= 18 + 20
= 48 m / s
v(av) = 38 + 48/ 2
= 43 m/swc
dv / dt = 10
a = 10 m/ s
Posted by Aman Saifi 5 years ago (7633557)
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Posted by Rajni Maheshwari 5 years ago (4191060)
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Posted by Avinash Kumar 5 years ago (10495461)
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Ronak Mor 4 years, 11 months ago (10319471)
Manshi Rawat 5 years ago (10506151)
Posted by Arushi Patidar 5 years ago (10135459)
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Posted by Sana Ms 4 years, 7 months ago (9846597)
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Sia ? 4 years, 7 months ago (6945213)
Posted by Vedha ..... 4 years, 9 months ago (9408112)
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Sia ? 4 years, 9 months ago (6945213)
WHEN THE LIFT IS ACCELERATING UPWARD :
let the acceleration of lift be a
Reaction force=R=m(g+a)
Hence apparent weight of the a man is more than actual weight.
WHEN THE LIFT IS ACCELERATING DOWNWARD
REACTION FORCE=m(g-a)
hence apparent weight of man is less than actual weight .
Posted by Sana Ms 5 years ago (9846597)
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Muskan Taneja 5 years ago (10507367)
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