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samigupta8
 3 years ago
find the magnitude of average accleration of the tip of second hand of length 10 cm during 10 seconds....
samigupta8
 3 years ago
find the magnitude of average accleration of the tip of second hand of length 10 cm during 10 seconds....

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0there are some assumptions which you should make is that, the seconds have move with constant angular velocity 'w' time period is 2pi/w=60. w=2pi/60 10 secs means it travels 1/6th part of the clock.dw:1382102831264:dw it goes from A to B. change in velocity and solve like the previous case how we did it. and then that divide by the total time taken=10secs. do this and you'll get the ans.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Radial acceleration is \(R\omega^2\) assuming constant angular velocity. The angular velocity is \(\dfrac{2\pi}{60} \rm rad s^{1}\)

samigupta8
 3 years ago
Best ResponseYou've already chosen the best response.0@AllTehMaffs pls..hlp me in this question

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ahhww.. i dunno how to even begin.. average acceleration shoudl be zero cause before 10 seconds and after 10 seconds, the second hand comes to stop :/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh wait.. are you talking about centripetal acceleration!?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then what parth says is right

samigupta8
 3 years ago
Best ResponseYou've already chosen the best response.0if it's right then how we r going to calculate acceleration

samigupta8
 3 years ago
Best ResponseYou've already chosen the best response.0i m convinced with this that linear velocity is cuming as pi/300 m/sec

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then u know what is the relation between linear vel and centripetal acceleration

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0there you ahve it.. plug it in!

samigupta8
 3 years ago
Best ResponseYou've already chosen the best response.0it's cuming as pi^2/90

samigupta8
 3 years ago
Best ResponseYou've already chosen the best response.0no no sory i made a mistake lemme solve it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0u knw u can solve this simply.. 60 seconds  2 pi rads 1 second  pi/30 rads so a = w^2 r = (pi^2/900)*0.1

samigupta8
 3 years ago
Best ResponseYou've already chosen the best response.0bt the answer is given as pi/3000 m/s^2

samigupta8
 3 years ago
Best ResponseYou've already chosen the best response.0here have they taken the value of one pi as 3 and then solve it

samigupta8
 3 years ago
Best ResponseYou've already chosen the best response.0if dey hve done like this then it wud give the same answer to our question also am i ryt....?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The answer is π²/9000

samigupta8
 3 years ago
Best ResponseYou've already chosen the best response.0ryt my answer is also that only

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That's also the answer Mashy gave you.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the answer given is correct. it's pi/3000 m/s^2. at least this is what i'm getting.

samigupta8
 3 years ago
Best ResponseYou've already chosen the best response.0how r u getting this answer cuz i m getting it as pi^2/9000

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hi @samigupta8, I am so sorry, but @Mashy and myself have led you to a wrong solution to your problem. Actually, @Abhishek619 gave you the right answer at the very beginning. As the problem asks for AVERAGE acceleration from position 1 to position 2, as acceleration is a vector, the right answer is: \(\vec a_{average} = \Large \frac {\vec v_2\vec v_1}{t_2t_1}\) with \(\Delta t = t_2t_1\)=10 s and \(\vec v_1\) and \(\vec v_2\) as drawn in Abhishek619's sketch. dw:1382428975483:dw Hence \(\vec v_2\vec v_1\) has the same magnitude v as v1 and v2 because all angles are π/3 the magnitude of the average acceleration is a = v/\(\Delta t\) Since v = 2πr/60s = 2π x 0.1 /60 = π / 300 m/s and \(\Delta t\) = 10 s then average acceleration is a = π / 3000 m/s²

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0drats.. thanks a lot for that update man!!
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