जब दो पिण्ड A व B, वेगो v_(A) व v_(B) से चल रहे हैं, तब A का B के सापेक्ष वेग v_(AB)=v_(A)-v_(B)...

Find V_(A)-V_(B)

V_A - V_B will be .

Two masses A and B are connected with two an inextesible string to write constraint relation between v_(A) & v_(B) Student A : v_(A) cos theta=v_(B) Student B : v_(B) cos theta=v_(A)

Planet A has massa M and radius R. Planet B has half the mass and half the radius of Planet A.If the escape velocities from the Planets A and B are v_(A) and v_(B), respectively , then (v_(A))/(v_(B))=n/4. The vlaue of n is :

omega_(AB)=("Relative velocity of A w.r.t. B perpendicular to line AB")/("Sepration between a and B") |(vec(v)_(AB))_(bot)|=v_(A) sin theta_(1)+v_(B) sin theta_(2) r_(AB)=r omega_(AB)=(v_(A) sin theta_(1)+v_(B) sin theta_(2))/(r) (v_(AB))_(11)=v_(A) cos theta_(1)-v_(B) cos theta_(2) (v_(AB))_(11) gt 0." " Then A will approach B (v_(AB))_(11) lt 0. " " Then B will separate from A

The circuit consists of resistors and ideal cells. I_(1) and I_(2) are current through branches indicated in the figure, V_(A) and V_(B) is the potential at points A and B on the circuit The value of V_(A) - V_(B) in volts is: