Because the tip of the moon's shadow ... the area of "totality" ... is never more than a couple hundred miles across, It never covers a single place for more than 7 minutes, and can never stay on the Earth's surface for more than a few hours altogether during one eclipse.
If you're not inside that small area, you don't see a total eclipse.
Answer:
The angular frequency of the block is ω = 5.64 rad/s
Explanation:
The speed of the block v = rω where r = amplitude of the oscillation and ω = angular frequency of the oscillation.
Now ω = v/r since v = speed of the block = 62 cm/s and r = the amplitude of the oscillation = 11 cm.
The angular frequency of the oscillation ω is
ω = v/r
ω = 62 cm/s ÷ 11 cm
ω = 5.64 rad/s
So, the angular frequency of the block is ω = 5.64 rad/s
As momentum / time = force
so; time = 100÷15
so your answer is 6.7 !!
Answer:
I = 8.75 kg m
Explanation:
This is a rotational movement exercise, let's start with kinetic energy
K = ½ I w²
They tell us that K = 330 J, let's find the angular velocity with kinematics
w² = w₀² + 2 α θ
as part of rest w₀ = 0
w = √ 2α θ
let's reduce the revolutions to the SI system
θ = 30.0 rev (2π rad / 1 rev) = 60π rad
let's calculate the angular velocity
w = √(2 0.200 60π)
w = 8.683 rad / s
we clear from the first equation
I = 2K / w²
let's calculate
I = 2 330 / 8,683²
I = 8.75 kg m
