According to the law , naturalized amrican citizens

2.60 m/s²

149 N and 137 N

Explanation:

Draw free body diagrams for the blocks and the pulley.

For the first block, there are two forces: m₁g pulling down and T₁ pulling up.

For the second block, there are two forces: m₂g pulling down and T₂ pulling up.

For the pulley, there are two torques: T₁r pulling counterclockwise and T₂r pulling clockwise.

m₁ moves up, so sum the forces in the +y direction:

∑F = ma

T₁ − m₁g = m₁a

m₂ moves down, so sum the forces in the -y direction:

∑F = ma

m₂g − T₂ = m₂a

The pulley rotates clockwise, so sum the torques in the clockwise direction:

∑τ = Iα

T₂r − T₁r = (½ Mr²) (a/r)

T₂ − T₁ = ½ Ma

If we add the first two equations together:

T₁ − m₁g + m₂g − T₂ = m₁a + m₂a

T₁ − T₂ + (m₂ − m₁) g = (m₁ + m₂) a

Substitute:

½ Ma + (m₂ − m₁) g = (m₁ + m₂) a

(m₂ − m₁) g = (m₁ + m₂ − ½ M) a

a = (m₂ − m₁) g / (m₁ + m₂ − ½ M)

Plug in values:

a = (19.0 kg − 12.0 kg) (9.8 m/s²) / (12.0 kg + 19.0 kg − ½ (9.20 kg))

a = 2.60 m/s²

Plug this into the first two equations to find the two tensions.

T₁ = m₁g + m₁a = 148.8 N

T₂ = m₂g − m₂a = 136.8 N

Rounded to three significant figures, the tensions are 149 N and 137 N.