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:

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) : .

Find the resultant of the following two vector vecA and vecB . vec(A):40 units due east and , vec(B):25 units 37^(@) north of west

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

Two forces vec(F)_(1)=500N due east and vec(F)_(2)=250N due north have their common initial point. vec(F)_(2)-vec(F)_(1) is

A smooth track is in the form of a quarter circle of radius 10 m lies in the vertical plane. A bead moves from A and B along the circular track under the action of forces vec(F)_(1),vec(F)_(2) and vec(F)_(3) Force vec(F)_(1) acts always towards point B and is always 20 N in magnitude Force vec(F)_(2) always acts horizontally and is always 25 N in magnitude. Force vec(F)_(3) always acts tangentially to the track and is of constant magnitude 10 N. Select the correct answer.

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 force F _(1), of magnitude 2.0N and directed due east is exerted on an object. A second force exerted on the object is F _(2)=2. 0 N, due north. Wha tis the magnitude and direction of a third force, F _(3), which must be exerted on the object so that the resultant force is zero ?

Two coaxial solenoids of different radius carry current I in the same direction. vec(F_1) be the magnetic force on the inner solenoid due to the outer one and vec(F_2) be the magnetic force on the outer solenide due to the inner one. Then

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 the resultant of vec(F_(1)) + vec(F_(2)) - vec(F_(3)) ( sin 37^(@) = (3)/(5) , cos 37^(@) = (4)/(5))