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Resultant force is zero.
R = (P2 + Q2 + 2PQ cos θ) ¹/² = (P² + Q² + 2P.P cos 2π/3)¹/² = [2P² + 2P² ( -1/2)] ¹/² = P
a₁/a₂ = tan30/tan60 = 1/√3 / √3 = 1/3 = 1:3
Use R= √(A^2+B^2+2 ABcosQ) R= √(3P^2+Q²) ,A=P=Q, B= P-Q Solve, θ=60° Establish the following vector inequalities :
a = dv/at = 3t²+2t+2)dt dv= (3t²+2t=2)dt ∫dv = ∫(3t^2+2t+2)dt v= t³+t²+2t+c c=2 m/s, v=18 m/s at t =2s.
What is the vector sum of n coplanar forces, each of magnitude F, if each force makes an angle 2π/n of with the preceding force?
Resultant force is zero.
Resultant force is zero.
See lessWhen the angle between two vectors of equal magnitudes is 2π/3, prove that the magnitude of the resultant is equal to either
R = (P2 + Q2 + 2PQ cos θ) ¹/² = (P² + Q² + 2P.P cos 2π/3)¹/² = [2P² + 2P² ( -1/2)] ¹/² = P
R = (P2 + Q2 + 2PQ cos θ) ¹/²
See less= (P² + Q² + 2P.P cos 2π/3)¹/²
= [2P² + 2P² ( -1/2)] ¹/² = P
The V-t graphs of two objects make angle 30° and 60° with the time axis. Find the ratio of their accelerations.
a₁/a₂ = tan30/tan60 = 1/√3 / √3 = 1/3 = 1:3
a₁/a₂ = tan30/tan60 = 1/√3 / √3 = 1/3 = 1:3
See lessAt what angle do the two forces (P + Q) and (P – Q) act so that the resultant is √(3P²+Q²).
Use R= √(A^2+B^2+2 ABcosQ) R= √(3P^2+Q²) ,A=P=Q, B= P-Q Solve, θ=60° Establish the following vector inequalities :
Use R= √(A^2+B^2+2 ABcosQ)
See lessR= √(3P^2+Q²) ,A=P=Q,
B= P-Q
Solve, θ=60°
Establish the following vector inequalities :
The acceleration of a particle in ms⁻² is given by a = 3t² + 2t + 2, where time t is in second. If the particle starts with a velocity v = 2 ms⁻¹ at t = 0, then find the velocity at the end of 2 s.
a = dv/at = 3t²+2t+2)dt dv= (3t²+2t=2)dt ∫dv = ∫(3t^2+2t+2)dt v= t³+t²+2t+c c=2 m/s, v=18 m/s at t =2s.
a = dv/at = 3t²+2t+2)dt
See lessdv= (3t²+2t=2)dt
∫dv = ∫(3t^2+2t+2)dt
v= t³+t²+2t+c
c=2 m/s, v=18 m/s at t =2s.