The object takes 0.5 seconds to complete one rotation, so its rotational speed is 1/0.5 rot/s = 2 rot/s.
Convert this to linear speed; for each rotation, the object travels a distance equal to the circumference of its path, or 2<em>π</em> (1.2 m) = 2.4<em>π</em> m ≈ 7.5 m, so that
2 rot/s = (2 rot/s) • (2.4<em>π</em> m/rot) = 4.8<em>π</em> m/s ≈ 15 m/s
thus giving it a centripetal acceleration of
<em>a</em> = (4.8<em>π</em> m/s)² / (1.2 m) ≈ 190 m/s².
Then the tension in the rope is
<em>T</em> = (50 kg) <em>a</em> ≈ 9500 N.
You draw a straight line from the start point to the end point. It doesn't matter what route was actually followed for the trip.
Answer:
Explanation:
Given that
Mass of bowling ball M1=7.2kg
The radius of bowling ball r1=0.11m
Mass of billiard ball M2=0.38kg
The radius of the Billiard ball r2=0.028m
Gravitational constant
G=6.67×10^-11Nm²/kg²
The magnitude of their distance apart is given as
r=r1+r2
r=0.028+0.11
r=0.138m
Then, gravitational force is given as
F=GM1M2/r²
F=6.67×10^-11×7.2×0.38/0.138²
F=9.58×10^-9N
The force of attraction between the two balls is
F=9.58×10^-9N
B. Exfoliation. Hope I helped you out bro.