A ball on the end of a string is whirled around in a horizontal circle of radius 0.478 m. The plane of the circle is 1.02 m above the ground. The string breaks and the ball lands 2.5 m away from the point on the ground directly beneath the ball’s location when the string breaks. The acceleration of gravity is 9.8 m/s 2 . Find the centripetal acceleration of the ball during its circular motion. Answer in units of m/s 2 .

Explanation:

We shall find first the velocity of ball at the time when string breaks. Let it be v . During its fall on the ground , 1.02 m below, we use the formula

h = 1/2 gt² where t is time of fall .

1.02 = 1/2 x 9.8 x t²

t²= .2081

t = .456

During this time it travels horizontally at distance of 2.5 m with uniform velocity of v

v x .456 = 2.5

v  = 5.48 m /s

centripetal acceleration

= v² / r where r is radius of the circular path

= 5.48² / .478

= 62.82 m /s²

colorectal cancer may also develop from areas of abnormal cells in the lining of the colon or rectum. this area of abnormal cells is called dysplasia and is more commonly seen in people with certain inflammatory diseases of the bowel such as crohn's disease or ulcerative colitis.

the inner core.

good luck: )

klicker