The moment of inertia is the rotational analog of mass, and it is given by
the product of mass and the square of the distance from the axis.
- The moment of inertia changes as the position of the weight is changed, which indicates that; statement is incorrect
Reasons:
The weight on each arm that have adjustable positions can be considered as point masses.
The moment of inertia of a point mass is <em>I</em> = m·r²
Where;
m = The mass of the weight
r = The distance (position) from the center to which the weight is adjusted
Therefore;
The moment of inertia, <em>I </em>∝ r²
Which gives;
Doubling the distance from the center of rotation, increases the moment of inertia by factor of 4.
Therefore, the statement contradicts the relationship between the radius of rotation and moment of inertia.
Learn more about moment of inertia here:
brainly.com/question/4454769
Answer:
Explanation:
a ) V = 3 cos(0.5t)
differentiating with respect to t
dv /dt = -3 x .5 sin0.5t
= -1.5 sin0.5t.
acceleration = - 1.5 sin 0.5t
when t = 3 s
acceleration = - 1.5 sin 1.5
= - 1.496 ms⁻²
v = 3 cos.5t
b ) dx/dt = 3 cos 0.5 t
dx = 3 cos 0.5 t dt
integrating on both sides
x = 3 sin .5t / .5
x = 6 sin0.5t
At t = 2 s
x = 6 sin 1
x = 5.05 m
Answer:
400 g
Explanation:
The computation of the number of grams in the original sample is shown below:
Given that
half-life = 5.26 years
total time of decay = 15.8 years
final amount = 50.0 g
Now based on the above information
number of half-lives past is
= 15.8 ÷ 5.26
= 3 half-lives
Now
3 half-lives = 1 ÷ 8 remains = 50.0 g
So, the number of grams would be
= 50.0 g × 8
= 400 g
Answer:
1, When Jane brakes, the brakes slow the car wheels turning and the road surface exerts a backwards force on the tires, causing the car to decelerate. The pocket book tends to continue on in a straight line (Newton's first law). If she brakes hard enough that the friction between the book and the car seat is insufficient to decelerate the book as fast as the car is decelerating, the book will slide off the seat, and gravity pulls it to the floor
2.
When the diver uses his / her force to depress the springboard, the springboard pushes him back with equal force
3.Newton's Second Law (F=ma)
4. 5 N
5. 19.5 N
65kg * 0.3 m/s^2
6.0.2 N/s
10kg divided by 2N
7.-Walking then pushing the moving forward
-Dribbling
-Basketball is pushed but bounces back
Explanation:
