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Record in the givens: OA=1760 m=1100 kg u=0 m/s SinѲ=opposite/hypotenuse=160/1760=1/11 K.E=1/2mv^2=1/2(1100)(v^2-0^2)=550v^2http://teachers.cie.org.uk/docs/dynamic/72493.pdf
question 5 help plz.....
Thanks Alot Brother...Record in the givens: OA=1760 m=1100 kg u=0 m/s SinѲ=opposite/hypotenuse=160/1760=1/11 K.E=1/2mv^2=1/2(1100)(v^2-0^2)=550v^2
Gain in P.E=mgh*sinѲ=1100*10(x-0)*1/11=1000x
Work done by driving force=gain in P.E + gain in K.E+Work done against resistance
1800x=1000x+550x^2+700x
100x=55ov^2
kv^2=x
55ok=100
k=5.5
In A.B there is no increase in height so Gain in P.E equals zero
Driving Force-Resistance=mass*acceleration
1800 – 700 = 1100a
a=1m/s^2
i)
Thank Broi)
using newtons 2nd law of motion
at A :
F=ma
Tension- weight=ma
T - 2.5 = 0.25a ---------equation 1
at B :
F=ma
weight-tension =ma (in this we do weight - tension coz the motion of the ball is in direction of weight so weight is the driving force)
7.5 - T = 0.75a --------equation 2
solve equation 1 and 2 simultaneously
u will obtain a=5 ms^-2<-------------
using kinematics formula s=ut = 0.5 * a * t^2
u=0 (strts from rest)
t=0.6
s= 0 + 0.5 *5 * (0.6) ^2
s= 0.9 m
sisThank Bro
Oh Sis sorry
i thought my gender was specified on my profile summary hihiOh Sis sorry
AB is perpendicular to OAB. hence the normal to AB lies on OAB as a direction.
AB is perpendicular to OAB. hence the normal to AB lies on OAB as a direction.
find n hat using cross product taking directions 1. the normal (OP) 2. direction vector of plane OAB
replace this in formula r.n hat = a. n hat
where a can be any point on plane. use any u want.
ii)View attachment 48106
Can anyone please help me out with the (ii) and (iii) part .
Thankyou in advance
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