A0.55-kg ball, attached to the end of a horizontal cord, is revolved in a circle of radius 1.3 m on a frictionless horizontal surface. if the cord will break when the tension in it exceeds 75 n, what is the maximum speed the ball can have?

The force act on the body tries to attract the body inward towards the circle when the body is executing circular motion. The maximum value of speed will be 13.3 m/sec.

The force operating on an object in curvilinear motion directed toward the axis of rotation or center of curvature is known as centripetal force.

Newton is the unit of centripetal force.

The centripetal force is always perpendicular to the direction of displacement of the item. The centripetal force of an item traveling on a circular route always works towards the center of the circle.

Due to rotation, the tension is act in the chord which is equal to the centripetal force act the ball.

Tension in the cord  = T= 75 N

Centripetal force =

Mass of the ball= m= 0.55 Kg

Linear  speed =v= ?

The radius of the circle.= r = 1.3 m

T is the maximum tension before the cord breaks. So according to the condition the obtained velocity will be ;

Hence the maximum value of speed will be 13.3 m/sec.

To learn more about the centripetal force refer to the link;

link

i think the answer would be reserve.