Answer:
I do agree with the student. Because imagene you have a tissue and a stick. And you throw them at the same speed , the stick will go farther because it' s heavier but the tissue would just blow around and not go forward because it is not heavy enought.
Explanation:
Hope this is correct have a terrific day.
Answer:
a) 
b)
Explanation:
Given that:
- mass of rod,
- length of the rod,
<u>(a)</u>
<u>Moment of inertia of rod about its center and perpendicular to the rod is given as:</u>
(b)
<u>Moment of inertia on bending the rod to V-shape of 60 degree angle and axis being perpendicular to the plane of V at the vertex.</u>
<em>We treat it as two rod with axis of rotation at the end and perpendicular to the plane of rotation. </em>
<em>So, the mass and the length of the rod will become half of initial value.</em>
![I=2[ \frac{1}{3}\times 0.2\times 0.3^2]](https://tex.z-dn.net/?f=I%3D2%5B%20%5Cfrac%7B1%7D%7B3%7D%5Ctimes%200.2%5Ctimes%200.3%5E2%5D)
Copper is the higher concentrated metal in this case.
<span>Answer:
First we need to find the acceleration.
torque on cylinder Ď„ = T * r where T is the string tension;
T = m(g - a) where a is the acceleration of the cylinder. Then
Ď„ = m(g - a)r
But also τ = Iα. For a solid cylinder, I = ½mr²,
and if the string doesn't slip, then α = a / r, so
τ = ½mr² * a/r = ½mra.
Since Ď„ = Ď„, we have
m(g - a)r = ½mra → m, r cancel, leaving
g - a = ½a
g = 3a/2
a = 2g/3 where g, of course, is gravitational acceleration.
We know that v(t) = a*t, so for our cylinder
v(t) = 2gt / 3 â—„ linear velocity
and ω = v(t) / r = 2gt / 3r ◄ angular velocity</span>
