Question

What will be kinetic energy of a body if it's mass is doubled.

Answer 1

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Kinetic energy of a body of given mass, is directly proportiional to its square of its velocity. When momentum is doubled, mass remaining constant, this means velcity is doubled. Hence, kinetic energy becomes 22=4 times i.e. say from 100 units to 400 units. and percentage increase in kinetic energy is 400−100=300% .

Answer 2

Kinetic Energy = (1/2) x (mass) x ((velocity)^2)

Kinetic Energy = (1/2) x (mass) x ((velocity)^2)mass - kg, velocity - m/s, KE - kJ

Kinetic Energy = (1/2) x (mass) x ((velocity)^2)mass - kg, velocity - m/s, KE - kJ(1) KE = (1/2)*(m)*(v^2)

Kinetic Energy = (1/2) x (mass) x ((velocity)^2)mass - kg, velocity - m/s, KE - kJ(1) KE = (1/2)*(m)*(v^2)So a doubling of mass, having a 1:1 ratio with KE, would double the kinetic energy. Doubling the velocity, (2v^2) = 4 times the KE.

Kinetic Energy = (1/2) x (mass) x ((velocity)^2)mass - kg, velocity - m/s, KE - kJ(1) KE = (1/2)*(m)*(v^2)So a doubling of mass, having a 1:1 ratio with KE, would double the kinetic energy. Doubling the velocity, (2v^2) = 4 times the KE.Doubling both would then produce KE = (1/2) x (2m) x ((2v)^2) .

Kinetic Energy = (1/2) x (mass) x ((velocity)^2)mass - kg, velocity - m/s, KE - kJ(1) KE = (1/2)*(m)*(v^2)So a doubling of mass, having a 1:1 ratio with KE, would double the kinetic energy. Doubling the velocity, (2v^2) = 4 times the KE.Doubling both would then produce KE = (1/2) x (2m) x ((2v)^2) .So in comparison to the original formula (1), and moving the multipliers to the front of the equation shows KE = (1/2) x (2)m x (4)(v^2).. .

Kinetic Energy = (1/2) x (mass) x ((velocity)^2)mass - kg, velocity - m/s, KE - kJ(1) KE = (1/2)*(m)*(v^2)So a doubling of mass, having a 1:1 ratio with KE, would double the kinetic energy. Doubling the velocity, (2v^2) = 4 times the KE.Doubling both would then produce KE = (1/2) x (2m) x ((2v)^2) .So in comparison to the original formula (1), and moving the multipliers to the front of the equation shows KE = (1/2) x (2)m x (4)(v^2).. .(2)*(4)*(1/2)*m*(v^2) = (8) * (1 x KE)

Kinetic Energy = (1/2) x (mass) x ((velocity)^2)mass - kg, velocity - m/s, KE - kJ(1) KE = (1/2)*(m)*(v^2)So a doubling of mass, having a 1:1 ratio with KE, would double the kinetic energy. Doubling the velocity, (2v^2) = 4 times the KE.Doubling both would then produce KE = (1/2) x (2m) x ((2v)^2) .So in comparison to the original formula (1), and moving the multipliers to the front of the equation shows KE = (1/2) x (2)m x (4)(v^2).. .(2)*(4)*(1/2)*m*(v^2) = (8) * (1 x KE)So, double the mass = 2 x KE

Kinetic Energy = (1/2) x (mass) x ((velocity)^2)mass - kg, velocity - m/s, KE - kJ(1) KE = (1/2)*(m)*(v^2)So a doubling of mass, having a 1:1 ratio with KE, would double the kinetic energy. Doubling the velocity, (2v^2) = 4 times the KE.Doubling both would then produce KE = (1/2) x (2m) x ((2v)^2) .So in comparison to the original formula (1), and moving the multipliers to the front of the equation shows KE = (1/2) x (2)m x (4)(v^2).. .(2)*(4)*(1/2)*m*(v^2) = (8) * (1 x KE)So, double the mass = 2 x KEdouble the velocity = 4 x KE

Kinetic Energy = (1/2) x (mass) x ((velocity)^2)mass - kg, velocity - m/s, KE - kJ(1) KE = (1/2)*(m)*(v^2)So a doubling of mass, having a 1:1 ratio with KE, would double the kinetic energy. Doubling the velocity, (2v^2) = 4 times the KE.Doubling both would then produce KE = (1/2) x (2m) x ((2v)^2) .So in comparison to the original formula (1), and moving the multipliers to the front of the equation shows KE = (1/2) x (2)m x (4)(v^2).. .(2)*(4)*(1/2)*m*(v^2) = (8) * (1 x KE)So, double the mass = 2 x KEdouble the velocity = 4 x KEdouble both = 8 x KE