What is the total kinetic energy of a 0. 15 kg hockey puck sliding at 0. 5 m/s and rotating about its center at 8. 4 rad/s? the diameter of the hockey puck is 0. 076 m.

The mass of the puck is

m = 0.15 kg.

The diameter of the puck is 0.076 m, therefore its radius  is

r = 0.076/2 = 0.038 m

The sliding speed is

v = 0.5 m/s

The angular velocity is

ω = 8.4 rad/s

The rotational moment of inertia of the puck is

I = (mr²)/2

 = 0.5*(0.15 kg)*(0.038 m)²

 = 1.083 x 10⁻⁴ kg-m²

The kinetic energy of the puck is the sum of the translational and rotational kinetic energy.

The translational KE is

KE₁ = (1/2)*m*v²

      = 0.5*(0.15 kg)*(0.5 m/s)²

      = 0.0187 j

The rotational KE is

KE₂ = (1/2)*I*ω²

      = 0.5*(1.083 x 10⁻⁴ kg-m²)*(8.4 rad/s)²

      = 0.0038 J

The total KE is

KE = 0.0187 + 0.0038 = 0.0226 J

0.0226 J

answer:

9+10= 21

explanation:

a

explanation:

csc