If the object is not at rest how?
Answer:
0.366×10^{-3} / s
Explanation:
θ = θmax e^{-bt/2m}
Given that
θ = 5.50°
θmax = 15.0°
So that we have
ln (θ / θmax) = -bt /2m
= - ln(5.50°/ 15.0°) / 1000s = b /2m
= b / 2m = 0.366×10^{-3} / s
Answer:
Explanation:
Initial angular velocity ω₀ = 151 x 2π / 60
= 15.8 rad /s
final velocity = 0
Angular deceleration α = 2.23 rad / s
ω² = ω₀² - 2 α θ
0 = 15.8² - 2 x 2.23 θ
= 55.99 rad
one revolution = 2π radian
55.99 radian = 55.99 / 2 π no of terns
= 9 approx .
∑F = ma = (90 kg)(1.2 m/s²) = 108 N = 100 N (1 significant digit)
Explanation:
We have,
Semimajor axis is 
It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :
G is universal gravitational constant
M is solar mass
Plugging all the values,
Since,
So, the orbital period of a dwarf planet is 138.52 years.
