(a) The amount of work required to change the rotational rate is 0.0112 J.
(b) The decrease in the rotational inertia when the outermost particle is removed is 64.29%.
<h3>
Moment of inertia of the rod</h3>
The moment of inertia of the rod from the axis of rotation is calculated as follows;
I = md² + m(2d)² + m(3d)²
where;
- m is mass = 10 g = 0.01 kg
- d = 3 equal division of the length
d = 6/3 = 2 cm = 0.02 m
I = md²(1 + 2² + 3²)
I = 14md²
I = 14(0.01)(0.02)²
I = 5.6 x 10⁻⁵ kg/m³
<h3>Work done to change the rotational rate</h3>
K.E = ¹/₂Iω²
K.E = ¹/₂(5.6 x 10⁻⁵)(60 - 40)²
K.E = ¹/₂(5.6 x 10⁻⁵)(20)²
K.E = 0.0112 J
<h3>Percentage decrease of rotational inertia when the outermost particle is removed</h3>
I₂ = md² + m(2d)²
I₂ = 5md²
ΔI = 14md² - 5md²
ΔI = 9md²
η = (ΔI/I) x 100%
η = (9md²/14md²) x 100%
η = 64.29 %
Learn more about rotational inertia here: brainly.com/question/14001220
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The Coulomb force is equal to the constant k times the product of charge one and charge two over radius.
F=k((q1q2)/r)
It transmits threw the neurons
Answer:
They use noise control, creating a wave that negates outside or ambient sound and replaces it with the desired sound that listeners request.
Explanation:
I hope this helped
Answer:
The final charges of each sphere are: q_A = 3/8 Q
, q_B = 3/8 Q
, q_C = 3/4 Q
Explanation:
This problem asks for the final charge of each sphere, for this we must use that the charge is distributed evenly over a metal surface.
Let's start Sphere A makes contact with sphere B, whereby each one ends with half of the initial charge, at this point
q_A = Q / 2
q_B = Q / 2
Now sphere A touches sphere C, ending with half the charge
q_A = ½ (Q / 2) = ¼ Q
q_B = ¼ Q
Now the sphere A that has Q / 4 of the initial charge is put in contact with the sphere B that has Q / 2 of the initial charge, the total charge is the sum of the charge
q = Q / 4 + Q / 2 = ¾ Q
This is the charge distributed between the two spheres, sphere A is 3/8 Q and sphere B is 3/8 Q
q_A = 3/8 Q
q_B = 3/8 Q
The final charges of each sphere are:
q_A = 3/8 Q
q_B = 3/8 Q
q_C = 3/4 Q
