If mass A is 1.0 kg, mass B is 5.0 kg, and the frictional pulley has a mass of 0.5 kg and a radius of 0.15 m, what is the velocity of A and B at 0.5 seconds? a. A = 1.72 m/s, B = 0.22 m/s
c. A = 0.22 m/s, B = 1.72 m/s
b. A = 3.14 m/s, B = -3.14 m/s
d. A = -3.14 m/s, B = 3.14 m/s

b. A = 3.14 m/s, B = -3.14 m/s

Explanation:

Assuming this refers to an Atwood machine, draw free body diagrams for each mass.

For mass A, there are two forces: tension T₁ pulling up and weight m₁g pulling down.

For mass B, there are two forces: tension T₂ pulling up and weight m₂g pulling down.

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

Sum of forces on A in the +y direction:

∑F = ma

T₁ − m₁g = m₁a

T₁ = m₁g + m₁a

Sum of forces on B in the -y direction:

∑F = ma

m₂g − T₂ = m₂a

T₂ = m₂g − m₂a

Sum of torques on the pulley in the clockwise direction:

∑τ = Iα

T₂r − T₁r = (½ mr²) (a/r)

T₂ − T₁ = ½ ma

Substitute:

m₂g − m₂a − (m₁g + m₁a) = ½ ma

m₂g − m₂a − m₁g − m₁a = ½ ma

m₂g − m₁g = m₂a + m₁a + ½ ma

g (m₂ − m₁) = a (m₂ + m₁ + ½ m)

a = g (m₂ − m₁) / (m₂ + m₁ + ½ m)

a = (9.8 m/s²) (5.0 kg − 1.0 kg) / (5.0 kg + 1.0 kg + ½ (0.5 kg))

a = 6.272 m/s²

Given:

v₀ = 0 m/s

a = 6.272 m/s²

t = 0.5 s

Find: v

v = at + v₀

v = (6.272 m/s²) (0.5 s) + 0 m/s

v = 3.14 m/s

Therefore, mass A will rise at 3.14 m/s, and mass B will fall at 3.14 m/s.

a high pressure system is a dense air mass that is usually cooler and drier than the surrounding air. a low pressure system is a less dense air mass that is usually wetter and warmer than the surrounding air.

in general, areas that experience high atmospheric pressure also experience fair weather. low pressure systems can cause the formation of clouds and storms. air usually flows from areas of high pressure to areas of low pressure..

think its the third option