A. a child is twirling a 1.52 kg object in a vertical circle with a radius of 67.6
cm at constant speed. the tension in the string is 54.7 n at the bottom of
the circle. (a.) what is the speed of the object?
b. (b) what is the tension in the string at the top of the circle?

(a) 4.21 m/s

(b) 24.9 N

Explanation:

(a) Draw a free body diagram of the object when it is at the bottom of the circle.  There are two forces on the object: tension force T pulling up and weight force mg pulling down.

Sum the forces in the radial (+y) direction:

∑F = ma

T − mg = m v² / r

v = √(r (T − mg) / m)

v = √(0.676 m (54.7 N − 1.52 kg × 9.8 m/s²) / 1.52 kg)

v = 4.21 m/s

(b) Draw a free body diagram of the object when it is at the top of the circle.  There are two forces on the object: tension force T pulling down and weight force mg pulling down.

Sum the forces in the radial (-y) direction:

∑F = ma

T + mg = m v² / r

T = m v² / r − mg

T = (1.52 kg) (4.21 m/s)² / (0.676 m) −  (1.52 kg) (9.8 m/s²)

T = 24.9 N

the answer is kevlar

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle