A 0.2-kg stone is attached to a string and swung in a circle of radius 0.6 m on a horizontal and frictionless surface. If the stone makes 150 revolutions per minute, the tension force of the string on the stone is:. a. 0.75 N.
b. 1.96 N.
c. 0.03 N.
d. 30 N.
e. 0.2 N.

the tension force of the string on the stone is 30 N

Option d) 30 N is the correct answer.

Explanation:

Given the data in the question;

mass m = 0.2 kg

radius r = 0.6 m

θ = 150 revolutions = 300π rad

time t = 60 seconds

we know that; Angular speed ω = θ / t

we substitute

ω = 300π / 60

ω = 5π rad

Linear speed of stone u = ω × r

we substitute

u = 5π × 0.6

u = 3π m/s

The tension force of the string on the stone is equal to centripetal force, which aid it move in circle;

so

T = mv² / r

we substitute

T = [ 0.2 × (3π)² ] / 0.6

T = 17.7652879 / 0.6

T = 29.6 ≈ 30 N

Therefore, the tension force of the string on the stone is 30 N

Option d) 30 N is the correct answer.

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j

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