If `|vec F_(1)timesvec F_(2)|`=`vec F_(1).vec F_(2)`, then `vec F_(1) + vec F_(2)`

A particle,in equilibrium,is subjected to four forces vec F_(1),vec F_(2),vec F_(3) and vec F_(4)vec F_(1)=-10hat k,vec F_(2)=u((4)/(13)hat i-(12)/(13)hat j+(3)/(13)hat k),vec F_(3)=v(-(4)/(13)hat i-(12)/(13)hat j+(3)/(13)hat k),vec F_(4)=w(cos thetahat i+sin thetahat j) then find the values of u,v and w

A smooth track in the form of a quarter circle of radius 6 m lies in the vertical plane. A particle moves from P_(1) to P_(2) under the action of forces vec F_(1),vecF_(2) and vec F_(3) Force vec F_(1) is 20N, Force vec F_(2) is always 30 N in magnitude. Force vec F_(3) always acts tangentiallly to the track and is of magnitude 15 N . Select the correct alternative (s) : .

The base vectors vec a_ (1), vec a_ (2), vec a_ (3) are given in terms vectors vec b_ (1), vec b_ (2), vec b_ (3) vec a_ (1) = 2vec b_ (1) + 3vec b_ (1) -vec b_ (1), vec a_ (2) = vec a_ (1) -2vec b_ (2) + 2vec b_ (3), vec a_ (3) = - 2vec b_ ( 1) + vec b_ (2) -2vec b_ (3) if vec F_ (1) = 3vec b_ (1) -vec b_ (2) + 2vec b_ (3), Express vec F in terms of vec a_ (1) , vec a_ (2), vec a_ (3)

A moving particle of mass m is acted upon by five forces vec(F)_(1),vec(F)_(2),vec_(F)_(3),vec(F)_(4) and vec(F)_(5 . Forces vec(F)_(2) and vec(F)_(3) are conservative and their potential energy functions are U and W respectively. Speed of the particle changes from V_(a) to V_b when it moves from position a to b. Which of the following statement is/are true – (a) Sum of work done by vec(F)_(1),vec(F)_(4) and vec(F)_(5)=U_(b)-U_(a)+W_(b)-W_(a) (b) Sum of work done by vec(F)_(1),vec(F)_(4) and vec(F)_(5)=U_(b)-U_(a)+W_(b)-W_(a)+1/2m(V_(a)^(2)-V_(a)^(2)) (c) Sum of work done by all five forces =1/2m(V_(b)^(2)-V_(a)^(2)) (d) Sum of work done by vec(F)_(2) and vec(F)_(3)=(U_(b)+W_(b))-(U_(a)+W_(a))

the base vectors vec a_(1),vec a_(2),vec longrightarrow3 are given in terms of base vectors vec b_(1),vec b_(2),vec b_(3) as vec a_(1)=2vec b_(1)+3vec b_(2)-vec b_(2)-vec b_(3)+vec a_(2)=vec b_(1)-2vec b_(2)+2vec b_(3) and vec a_(3)=-2vec b_(1)+2vec b_(2)-2vec b_(3) If vec F=3vec b_(1)-vec b_(2)+2vec b_(3), then express vec F in terms of vec a_(1),vec a_(2) and vec a_(3)

vec(i)timesvec(j)= (a) vec0 (b) 1 (c) -vec k (d) vec k

Five forces vec(F)_(1) ,vec(F)_(2) , vec(F)_(3) , vec(F)_(4) , and vec(F)_(5) , are acting on a particle of mass 2.0kg so that is moving with 4m//s^(2) in east direction. If vec(F)_(1) force is removed, then the acceleration becomes 7m//s^(2) in north, then the acceleration of the block if only vec(F)_(1) is action will be:

Find magnitude and direction of vec F_(2)-vec F_(1) if vec F_(1)=500N due east and vec F_(2)=250N due north:

Consider a tetrahedron with faces f_(1),f_(2),f_(3),f_(4). Let vec a_(1),vec a_(2),vec a_(3),vec a_(4) be the vectors whose magnitudes arerespectively equal to the areas of f_(1),f_(2),f_(3),f_(4) and whose directions are prpendicular to these faces in theoutward direction.Then

When force vec(F_(1)), vec(F_(2)),vec(F_(3))"…..."vec(F_(n)) act on a particle , the particle remains in equilibrium . If vec( F_(1)) is now removed then acceleration of the particle is