Answer:
Explanation:
Given that,
Angular speed of a skater, 
The moment of inertia of the skater, I = 0.6 kg-m²
We need to find the angular momentum of the skater. The formula for the angular momentum of the skater is given by :
Substitute all the values,
So, its angular momentum is equal to 
.
Answer:
If a vertical line extending down from an object's CG extends outside its area of support, the object will topple
Explanation:
We can understand better this situation using a diagram with the forces acting on it.
In the attached image we can see that when the gravity center is bouncing outside from the area of the pedestal, the object will be out of balance and will fall.
The answer is parallel
If the <span>circuits in a car</span> were series, they would go out at the same time.
I hope this helps! :3
complete question:
A child bounces a 60 g superball on the sidewalk. The velocity change of the superball is from 22 m/s downward to 15 m/s upward. If the contact time with the sidewalk is 1/800 s, what is the magnitude of the average force exerted on the superball by the sidewalk
Answer:
F = 1776 N
Explanation:
mass of ball = 60 g = 0.06 kg
velocity of downward direction = 22 m/s = v1
velocity of upward direction = 15 m/s = v2
Δt = 1/800 = 0.00125 s
Linear momentum of a particle with mass and velocity is the product of the mass and it velocity.
p = mv
When a particle move freely and interact with another system within a period of time and again move freely like in this scenario it has a definite change in momentum. This change is defined as Impulse .
I = pf − pi = ∆p
F = ∆p/∆t = I/∆t
let the upward velocity be the positive
Δp = mv2 - m(-v1)
Δp = mv2 - m(-v1)
Δp = m (v2 + v1)
Δp = 0.06( 15 + 22)
Δp = 0.06(37)
Δp = 2.22 kg m/s
∆t = 0.00125
F = ∆p/∆t
F = 2.22/0.00125
F = 1776 N
