Thursday

November 27, 2014

November 27, 2014

Total # Posts: 117

**physics**

How the spring is connected to the block?
*June 18, 2012*

**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)
*June 18, 2012*

**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.
*June 18, 2012*

**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...
*June 18, 2012*

**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 ...
*June 18, 2012*

**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) + ...
*June 16, 2012*

**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
*June 14, 2012*

**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 ...
*June 14, 2012*

**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
*June 13, 2012*

**physics**

Momentum = mass * velocity You know two parameters of the above equation - so you can find the third.
*June 13, 2012*

**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 ...
*June 13, 2012*

**Physics**

Resistance R = r1* l/a1 = r2*l/a2 where r1 & r2 are resistivies of two wires. So a1/a2 = r1/r2
*June 13, 2012*

**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 - ...
*June 6, 2012*

**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...
*June 5, 2012*

**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)
*May 26, 2012*

**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^...
*May 26, 2012*

**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.
*May 23, 2012*