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Discussion Forum : Physics - Section 1 (Q.No. 60)
60.
ML2T-2 is the dimensional formula for
Discussion:
14 comments Page 1 of 2.
Asha Stephen said:
1 decade ago
Formula for Couple acting on body (T)=F*d
=Ma*d [F=M*a]
=M*dv/t*d [a==(vf-vi)/t)]
=M*(d/t/t)*d
=Md^2/t^2
=ML^2T^-2
=Ma*d [F=M*a]
=M*dv/t*d [a==(vf-vi)/t)]
=M*(d/t/t)*d
=Md^2/t^2
=ML^2T^-2
BALIRAM said:
1 decade ago
Two equal and opposite parallel forces acting along different lines on a body constitute a couple.
The product of force and perpendicular distance between two force.
So force*distance = couple of force.
UNIT: NEWTON*METER = MLT^-2 L = ML^2T^-2.
The product of force and perpendicular distance between two force.
So force*distance = couple of force.
UNIT: NEWTON*METER = MLT^-2 L = ML^2T^-2.
Venkatraj.A said:
1 decade ago
T = f.d.
Then T = ma.d.
= M.L/T2.L (a=L/T2) & (d=L).
= ML2/T2.
= ML2T-2.Formula for Couple acting on body (T)=F*d.
= Ma*d [F=M*a].
= M*dv/t*d [a==(vf-vi)/t)].
= M*(d/t/t)*d.
= Md^2/t^2.
= ML^2T^-2.
Then T = ma.d.
= M.L/T2.L (a=L/T2) & (d=L).
= ML2/T2.
= ML2T-2.Formula for Couple acting on body (T)=F*d.
= Ma*d [F=M*a].
= M*dv/t*d [a==(vf-vi)/t)].
= M*(d/t/t)*d.
= Md^2/t^2.
= ML^2T^-2.
Siddarrh said:
1 decade ago
Work and couple are two different things work helps us to move an object moment or couple indicates the possibilities for the work to be done.
Sujit said:
1 decade ago
COUPLE ACTING ON BODY MEANS TORQUE. SO USING THE FORMULA:
T = f.d.
THEN T= ma.d.
= M.L/T2.L (a=L/T2) & (d=L).
= ML2/T2.
= ML2T-2.
T = f.d.
THEN T= ma.d.
= M.L/T2.L (a=L/T2) & (d=L).
= ML2/T2.
= ML2T-2.
Rustam xaman said:
8 years ago
t=f.d f=ma m=M and d=L.
= M(a=v/t).L where v= s/t s=L
=M(L/t2).L WHERE t2 go square
=ML2T-2.
= M(a=v/t).L where v= s/t s=L
=M(L/t2).L WHERE t2 go square
=ML2T-2.
Ashish said:
6 years ago
Couple acting on body, unit is N-mm.
N = kg.mm.s^-2.
N-mm = kg.mm^2.s^-2.
N-mm = ML^2T^-2.
N = kg.mm.s^-2.
N-mm = kg.mm^2.s^-2.
N-mm = ML^2T^-2.
Rishi said:
1 decade ago
Force =m*a
and couple= force*distance
units of force =m*l*t^-2
and couple= ml^2t^-2.
and couple= force*distance
units of force =m*l*t^-2
and couple= ml^2t^-2.
Debjyoti Bandyopadhyay said:
1 decade ago
Couple = Force*distance between the force = Work.
Dimension of Work is ML^2T^-2.
Dimension of Work is ML^2T^-2.
Prathap said:
8 years ago
T = f.d.
Then T = ma.d.
= M.L/T2.L (a=L/T2) & (d=L).
= ML2/T2.
Then T = ma.d.
= M.L/T2.L (a=L/T2) & (d=L).
= ML2/T2.
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