Messages posted by nik_kaus

go to http://www.goiit.com/posts/list/mechanics-rotational-motion-23697.htm

go to http://www.goiit.com/posts/list/community-shelf-the-jee-guide-72019.htm

use u r lab manuals

wave length, (lambda) = vT = 10 * 2 * 10^–2 = 0.2 m = 20 cm

phase diff=(2pi/lambda*x) therfore=pi rad (v=10m/sec)

y₁ = a sin (wt – kx) implies 1.5 = a sin (wt – kx) (w=omega)

the displacement of the particle at a distance x = 10 cm

phi=2(pi)x/(lambda) = pi

y₂= a sin (wt – kx + pi) implies –a sin(wt – kx) = –1.5 mm

From the free body diagram

g = 10m/s², m = 2kg , $ = 30°, (nu) = 0.2 (nu)=coeff. of friction

R – mg cos $ - F sin $ = 0

R = mg cos $ + F sin $ ...(1)

And mg sin $ + (nu)R – F cos $ = 0

mg sin $ + nu(mg cos $ + F sin $) – F cos $ = 0

mg sin$ + (nu) mg cos $ + (nu) F sin $ – F cos $ = 0

F = mgsin$ - (nu)mgcos$/(nu)sin$ - cos$

F=2*10*0.5+0.2*2*10*(root3)/2/0.2*0.5 - (root3)/2

=17.7N

net unbalanced force cause shm x be displacement therfore

ma=pi(r)²x(ro)g (ro)=density of water

as t=(2pi)root(x/a) therefore (2pi)root(x*2*10³/pir²(ro)gx (ro)=density of water

therefore (2pi)root2*10³/pi*10²*1*10

therefore 2(pi)root2/pi*10

therefore .5sec

37.Suppose the monkey accelerates upward with acceleration ’a’& the block, accelerate downward with

     acceleration a₁. Let Force exerted by monkey is equal to ‘T’

     From the free body diagram of monkey

     T – mg – ma = 0 ...(i)

     T = mg + ma.

     Again, from the FBD of the block,

     T = ma₁ – mg = 0.

   mg + ma + ma₁ – mg = 0 [From (i)] ma = –ma₁ a = a_1.

    Acceleration ‘–a’ downward i.e. ‘a’ upward.

    The block & the monkey move in the same direction with equal acceleration.

    If initially they are rest (no force is exertied by monkey) no motion of monkey of block occurs as they

    have same weight (same mass). Their separation will not change as time passes

36. m = 15 kg of monkey, a = 1 m/s².

      From the free body diagram

T – [15g + 15(1)] = 0 , T = 15 (10 + 1) , T = 15 × 11 , T = 165 N.

      The monkey should apply 165N force to the rope.

      Initial velocity u = 0 ; acceleration a = 1m/s² ; s = 5m.

as s=ut+1/2at²

5=0+(1/2)1t²therefore t=root10

http://www.goiit.com/posts/list/mechanics-newtons-laws-of-motion-h-c-verma-47945.htm

refer to link http://www.goiit.com/posts/list/mechanics-hc-verma-pg-no-82-newton-s-laws-of-motion-exercise-59806.htm

Newton’s first law of motion can be interpreted to mean that objects do not change its state of motion on its own. This property of an object is known as “inertia” i.e. inactivity.

The property of the object of maintaining state of motion unless being forced externally is characterized by first law. For this reason, Newton’s first law is also referred as “law of inertia”, “inertia” is quantized by Newton’s second law of motion, which measures the “unwillingness” on the part of an object to change. Every object resists change in velocity.A body moves with acceleration only when it is acted by external net force. The amount of acceleration i.e. rate of change in velocity for a given net force is different for different mass of the bodies.From Newton's second law of motionF = ma. In words, a smaller mass yields a greater acceleration and a greater mass yields a smaller acceleration. Thus, “mass” of a body is the measure of the inertia of the body in translational motion. Here, we have specified that mass is the measure of inertia in translational motion, because the inertia to rotational motion is measured by a corresponding rotational term called "moment of inertia".