Answer:
When a ball on one end of the cradle is pulled away from the others and then released, it strikes the next ball in the cradle, which remains.
So it will be D
Explanation: Hope this helps :)
If the coordinates of the initial point and the end point of a vector is given, the Distance Formula can be used to find its magnitude. resultant force is the force (magnitude and direction) obtained when two or more forces are combined.
(a) The angular speed of the system at the instant the beads reach the end of the rod is 9.26 rad/s.
(b) The angular speed of the rod after the after the beads fly off the rod's ends is 25.71 rad/s.
<h3>Moment of inertia through the center of the rod</h3>
I = ¹/₁₂ML²
I = ¹/₁₂ (0.1)(0.5)²
I = 0.0021 kgm²
For the beads, I = 2Mr² = 2(0.03 x 0.1²) = 0.0006 kgm²
Total initial moment of inertia, Ii = 0.0021 kgm² + 0.0006 kgm²
I(i)= 0.0027 kgm²
When the beads reach the end, I = 2Mr² = 2(0.03)(0.25)² = 0.00373 kgm²
Total final moment of inertia, I(f) = 0.0021 kgm² + 0.00373 kgm²
I(f) = 0.00583 kgm²
<h3>Speed of the system</h3>
The speed of the system at the moment the beads reach the end of the rod is calculated as follows;
<h3>Speed of the rod when the beads fly off</h3>
Learn more about moment of inertia of rods here: brainly.com/question/3406242
Answer:
B. Is constantly present
Answer:
(a) α = -0.16 rad/s²
(b) t = 33.2 s
Explanation:
(a)
Applying 3rd equation of motion on the circular motion of the tire:
2αθ = ωf² - ωi²
where,
α = angular acceleration = ?
ωf = final angular velocity = 0 rad/s (tire finally stops)
ωi = initial angular velocity = 5.45 rad/s
θ = Angular Displacement = (14.4 rev)(2π rad/1 rev) = 28.8π rad
Therefore,
2(α)(28.8π rad) = (0 rad/s)² - (5.45 rad/s)²
α = -(29.7 rad²/s²)/(57.6π rad)
<u>α = -0.16 rad/s²</u>
<u>Negative sign shows deceleration</u>
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(b)
Now, we apply 1st equation of motion:
ωf = ωi + αt
0 rad/s = 5.45 rad/s + (-0.16 rad/s²)t
t = (5.45 rad/s)/(0.16 rad/s²)
<u>t = 33.2 s</u>
