a ball with a mass of 8.0kg is suspended by two wires: 1) 0.6 meters and 2) 1.0 meters from a horizontal arm of 0.8 meters, which is attached to a vertical shaft. The two wires and the horizontal arm forms a triangle. The shaft is in uniform rotation about its axis such that the linear speed of the ball equals 2.3 m/s. Find the tension of wire 1.

To find the tension of wire 1, we can use the centripetal force formula:

F = (m * v^2) / r

where:
F = Centripetal force
m = Mass of the ball (8.0 kg)
v = Linear speed of the ball (2.3 m/s)
r = Distance from the center of rotation (0.6 meters)

First, let's calculate the centripetal force on the ball:
F = (8.0 kg * (2.3 m/s)^2) / 0.6 meters

F = (8.0 kg * 5.29 m^2/s^2) / 0.6 meters

F = (42.32 kg * m^2/s^2) / 0.6 meters

F = 70.53 N

The centripetal force acting on the ball is 70.53 N.

Since wire 1 is responsible for providing this force, the tension in wire 1 will be equal to the centripetal force:

Tension of wire 1 = 70.53 N

Therefore, the tension in wire 1 is 70.53 N.