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) `

Calculate the threshold freqency of the metalfrom the photoelectronsare emitted with zero velocity when exposed to radiation of wavelength 6800 Å Hint : KE = (1)/(2) mv^(2) hv= hv_(0) + KE, KE=0 v= (c ) /( lambda)

Calculate the threshold freqency of the metal from the photoelectrons are emitted with zero velocity when exposed to radiation of wavelength 6800 Å Hint : KE = (1)/(2) mv^(2) hv= hv_(0) + KE, KE=0 v= (c ) /( lambda)

Calculate the threshold freqency of the metalfrom the photoelectronsare emitted with zero velocity when exposed to radiation of wavelength 6800 Å Hint : KE = (1)/(2) mv^(2) hv= hv_(0) + KE, KE=0 v= (c ) /( lambda)

The kinetic energy passessed by a body of mass m moving with a velocity v is equal to (1)/(2)mv^(2) , provided

Velocity of centre of mass of two particles is v and the sum of masses of two particles is m . Kinetic energy of the system (A)will be equal to (1)/(2) mv^(2) (B)will always be less than (1)/(2) mv^(2) (C)will be greater than or equal to (1)/(2) mv^(2) (D)will always be greater than (1)/(2) mv^(2)

State whether the following statements are true or false: The kinetic energy of a body of mass, , moving with velocity v is 1/2 mv^2 .

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