`E= 1/2 mv^2` સમીકરણમાંથી ઊર્જાના પરિમાણીય સૂત્રની ગણતરી કરો.

Calculate the frequency of paritcle wave, when the kinetic energy of a sub-atomic particle is 3.79 xx 10^(-27) J . Hint : KE = (1)/(2) mv^(2) lambda = (h)/(mv) , lambda = ( v)/( v) :. (v)/(v) = (h)/( mv), v= (mv^(2))/( h) = ( 2 xx ( 1)/(2) mv^(2))/( h)

For Question , calculat the block's speed asit passes through the equilibrium point ? Hint : E = K+U = (1)/(2) mv^(2) + (1)/(2) kx^(2) = (1)/(2) mv^(2) +0 . Calculate v

The equation for energy (E) of a simple harmonic oscillator, E = (1)/(2) mv^(2)+(1)/(2)+ omega^(2)x^(2) ?, is to be made "dimensionless" by multiplying by a suitable factor, which may involve the constants, m(mass), omega (angular frequency) and h(Planck's constant). What will be the unit of momentum and length?

Relation between kinetic energy and momentum Let us consider a body of mass 'm' having a velocity 'v', then momentum of the body P = mass xx velocity P = m xx v rArr v = (P)/(m) " "...(1) From definition, kinetic energy (K.E) of the body K.E = (1)/(2) mv^(2)" "...(2) Now putting the value of (1) in (2) we have K.E = (1)/(2)m ((P)/(m))^(2) K.E. = (1)/(2)m (P^(2))/(m^(2)) = (1)/(2) (P^(2))/(m) = (P^(2))/(2m)" "...(3) Thus we can write P^(2) = 2m xx K.E rArr P = sqrt(2m xx K.E) Thus momentum = sqrt(2 xx "mass" xx "kinetic energy") The kinetic energy of a given body is doubled. Its momentum will

A body whose mass is 3kg performs rectilinear motion according to the formula s = 1 + t + t^2 , where s is measured the kinetic energy 1/2 mv^2 and t in second. Determine the kinetic energy 1/2 mv^2 of the body in 5 sec after its start.

In photoelectric equation hv = hv _ 0 + (1 )/(2) mv ^ 2 of Einstein which classical law is followed

A meter stick of length l and mass M lies on a frictionlesss table. The stick is free to move in any way on the table. A hockey ball of mass m moving perpendicuarly to the mass of the ball so that it remains at rest immediately after the collision? [Hint: Elastic collision means that kinetic energy is conserved in the process. Consider conservation of kinetic energy, conservation of angualr momentum and linear momentum, i.e., 1/2mv^(2) = (1/2Mv^(2') + 1/2Iepsilon^(2) , mv1/2 = Iepsilon , mv = Mv^(') . Eliminate v^(2') and w^(2) from the first with the help of the other two. Use the expression I = 1/12 MI^(2)]

Assertion: If the intensity of incident radiation on a metal is doubled, then K.E. of photoelectrons emitted will be doubled. Reason: K.E.=1/2mv^2=h(v-v_0), where v is the velocity of photo electrons.

Given KE=(1)/(2)mv^(2) . If m=10kg, v=5m//s then find the value of KE?