सिद्ध कीजिए कि गतिज ऊर्जा `K.E.=(1)/(2)mv^(2).`

यदि triangleABC में (If इन triangleABC), a = (1)/(sqrt(6)-sqrt(2)), b = (1)/(sqrt(6)+sqrt(2)), C = 60^(@) तो सिद्ध कीजिए कि (Prove that) c = (sqrt(4))/(2) .

सिद्ध कीजिए कि 2tan^(-1)sqrt((b)/(a))=cos^(-1)((a-b)/(a+b))

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

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") Two bodies A and B of unequal masses having same momentum have masses in the ratio 1 : 2 then their K.E are in the ratio

सिद्ध कीजिए कि |vecA xx vecB|^(2) + |vecA * vecB|^(2) = (AB)^(2).

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)