Answer:
a) w = 4.24 rad / s
, b) α = 8.99 rad / s²
Explanation:
a) For this exercise we use the conservation of kinetic energy,
Initial. Vertical bar
Emo = U = m g h
Final. Just before touching the floor
Emf = K = ½ I w2
As there is no friction the mechanical energy is conserved
Emo = emf
mgh = ½ m w²
The moment of inertial of a point mass is
I = m L²
m g h = ½ (m L²) w²
w = √ 2gh / L²
The initial height h when the bar is vertical is equal to the length of the bar
h = L
w = √ 2g / L
Let's calculate
w = RA (2 9.8 / 1.09)
w = 4.24 rad / s
b) Let's use Newton's equation for rotational motion
τ = I α
F L = (m L²) α
The force applied is the weight of the object, which is at a distance L from the point of gro
mg L = m L² α
α = g / L
α = 9.8 / 1.09
α = 8.99 rad / s²
Answer:
Titan takes 11.634 times longer to orbit Saturn as compared to Enceladus.
Explanation:
We have been given that the average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km.
We will use Kepler's Law to solve our given problem.
Upon substituting our given values, we will get:
Taking square root of both sides, we will get:
This implies that time period of Titan about Sturn is 11.634 times more compared to time period of Enceladus about Saturn.
So, basically Titan takes 11.634 times longer to orbit Saturn as compared to Enceladus.
-- The mass of the sun never increases.
-- It does decrease, but not nearly enough to have any noticeable
effect on the orbital motion of the Earth, or any other planet.
-- When Earth is closer to the sun, it moves faster in its orbit.
-- When Earth is farther from the sun, it moves slower in its orbit.
-- The result is that the line from the sun to the Earth always covers
the same amount of area in the same length of time.
-- Johannes Kepler noticed this, and it's his Second Law of planetary motion.
-- Newton showed that if his equations for gravity and motion are correct,
then planets MUST behave this way.
Answer:
electrons
Explanation:
Given that,
Total charge = 9 mC = 0.009 C
0.009 C of charge passes through a wire in 3.6 s.
Let q' is the charge that passes through it in 10 s.
So,

We know that,
q = ne
Where
n is the number of electrons
So,

So,
electrons must pass through the cross-sectional area.