A disk of radius 10 cm is pulled along a frictionless surface with a force of 16 N by a string wrapped around the edge. At the i
1 answer:
Answer:
t = 0.2845Nm (rounded to 4 decimal places)
Explanation:
The disk rotates at a distance of an arc length of 28cm
Arc length = radius × central angle × π/180
28cm = 10cm × central angle × π/180
Central angle = × 180/π ≈ 160.4°
Torque (t) = rFsin(central angle) , where F is the applied force
Radius in meters = 10/100 = 0.1m
t = 0.1m × 16N × sin160.4°
t = 0.2845Nm (rounded to 4 decimal places)
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Answer:
103239.89 days
Explanation:
Kepler's third law states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.
a³ / T² = 7.496 × 10⁻⁶ (a.u.³/days²)
where,
a is the distance of the semi-major axis in a.u
T is the orbit time in days
Converting the mean distance of the new planet to astronomical unit (a.u.)
1 a.u = 9.296 × 10⁷ miles
Substituting the values into Kepler's third law equation;
(days)²
T = 103239.89 days
An estimate time T for the new planet to travel around the sun in an orbit is 103239.89 days