# Posts by ajayb

Total # Posts: 117

**physics**

How the spring is connected to the block?

**Physics**

Because Torque is a cross product of radius vector(r) and Force vector(F). If r =Ax+By+Cz & F = ax+by+cz (where x,y and z are unit vectors in x,y and z directions) then x-component of torque Tx is : Tx = (Bc-Cb) = 3*0-3*4 = -12 Therefore x-comp. of the Torque is nagative (-12)

**Physics**

Speed(of first wave) = freq*wavelength (m/s) Speed of the second wave is also the same. Since its wavelength is given, find the frequency. Reciprocal of the frequency is the time period in secs.

**Physics Help**

First compute mu(static): When the box just begins to move, the applied force equals mu(s)*m*g So mu(s)= F/m*g = 26.9/(9.1*9.8)= 0.30 Now when the box gets into motion, mu(k) comes in picture and frictional force f = mu(k)*m*g So F - mu(k)*m*g = m*a mu(k) = (F - m*a)/m*g = (26...

**physics**

Consider 'Free body diagram' of one of the beams (say left one) i.e. isolate it and see the forces acting on it. Let's say end B is on the ground and end A is hinged to the other beam of the frame. Which are the forces acting on it? 1.Normal reaction N (vertically ...

**Physics**

a) linear speed v = w*r (where w is the angular speed in radians/sec and r is the radial distance from the center) Here, w = 2990 rpm = 49.83 rev.per sec. = 49.83*2*pi rad/sec =313.11 rad/sec & r = 0.59/2 = 0.295 m So, v = 313.11*0.295 = 92.4m/s b) w(final) = w(initial) + ...

**physics....**

Since the post-lunch momentum is zero, the momentum of larger fish should be equal to the momentum of smaller fish in terms of magnitude but opposite in direction. Speed of the smaller fish v2: v2 = (m1*v1)/m2 = 5*1/1 = 5m/s

**physics....**

a) Stationary observer on the train: As the observer himself is having speed that of the train, Indiana Jones would appear to be running at 10 km/h. b) An observer on the train running in opp. direct to the train at speed of 10km/h: Relative to such an observer, speed of ...

**physics**

Momentum of the 'two boys' system shall be conserved (because there is no external force in the horizontal direction) So, m1v1 = m2v2 m1 = 40*10/8 = 50 Kg

**physics**

Momentum = mass * velocity You know two parameters of the above equation - so you can find the third.

**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 ...

**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 - ...

**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.