Physics
The charged particle on the incline stays in the middle because the electrstatic force of repulsion balances the gravitational force acting along the incline downwards. If the distance between the charged particles is x: k Q^2/x^2 = mg*(sin21deg) here k =1/4*pi*epsilon0 Comput...
Physics
Resistance R = r1* l/a1 = r2*l/a2 where r1 & r2 are resistivies of two wires. So a1/a2 = r1/r2
physics
If the amplitude is A, you are required to find the velocity of the mass when it is at a distance A/2 from the equilibrium position (i.e. midway between the extreme and equilibrium positions) use the expression: v = w*sqrt(A^2 - x^2) where v - velocity at displacement x x - di...
Physics
The momentum of the book is m1*v1 in +ve x-direction. Since there is no external force in the horizontal direction, the linear momentum of the 'book and you' system will be conserved. Therefore you would have the same magnitude of momentum in the opposite direction. If...
physics
X distance moved in first 2 secs. => x=0.2+(1/2).a.2^2= 2a ...(1) velocity at the end of 2 secs.(say v) => v = 0+a.2 = 2a ....(2) y distance covered in the next 2 secs. => y = v.t +(1/2).a.t^2 or y = 2a.2 + (1/2).a.2^2 = 6a ...(2) From, (1) & (2) y = 3x ...Option (2)
physics
Assume the particle's acceleration is a cm/sec^2 and that it moves a distance of S(t)cms in total time of t secs. Also it covers S(t-1) cms. in (t-1) secs. Since it covers 6cm in the first second => 6 = 0.1 + (1/2).a.1^2 => a=12cm/sec^2. Now, S(t)-S(t-1)= (1/2).a.[t^...
college physics
You are solving the problem correctly. Having found the acceleration of each block to be a= 2.4 mtr/sec sq, you need to feed its value in the second equation: T = m1a + m1g sin(theta1) and get the answer as T = 24.0 Nt.
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