(1) The linear acceleration of the yoyo is 3.21 m/s².
(2) The angular acceleration of the yoyo is 80.25 rad/s²
(3) The weight of the yoyo is 1.47 N
(4) The tension in the rope is 1.47 N.
(5) The angular speed of the yoyo is 71.385 rad/s.
<h3> Linear acceleration of the yoyo</h3>
The linear acceleration of the yoyo is calculated by applying the principle of conservation of angular momentum.
∑τ = Iα
rT - Rf = Iα
where;
- I is moment of inertia
- α is angular acceleration
- T is tension in the rope
- r is inner radius
- R is outer radius
- f is frictional force
rT - Rf = Iα ----- (1)
T - f = Ma -------- (2)
a = Rα
where;
- a is the linear acceleration of the yoyo
Torque equation for frictional force;

solve (1) and (2)

since the yoyo is pulled in vertical direction, T = mg 
<h3>Angular acceleration of the yoyo</h3>
α = a/R
α = 3.21/0.04
α = 80.25 rad/s²
<h3>Weight of the yoyo</h3>
W = mg
W = 0.15 x 9.8 = 1.47 N
<h3>Tension in the rope </h3>
T = mg = 1.47 N
<h3>Angular speed of the yoyo </h3>
v² = u² + 2as
v² = 0 + 2(3.21)(1.27)
v² = 8.1534
v = √8.1534
v = 2.855 m/s
ω = v/R
ω = 2.855/0.04
ω = 71.385 rad/s
Learn more about angular speed here: brainly.com/question/6860269
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Answer:
<em>Correct choice: b 4H</em>
Explanation:
<u>Conservation of the mechanical energy</u>
The mechanical energy is the sum of the gravitational potential energy GPE (U) and the kinetic energy KE (K):
E = U + K
The GPE is calculated as:
U = mgh
And the kinetic energy is:

Where:
m = mass of the object
g = gravitational acceleration
h = height of the object
v = speed at which the object moves
When the snowball is dropped from a height H, it has zero speed and therefore zero kinetic energy, thus the mechanical energy is:

When the snowball reaches the ground, the height is zero and the GPE is also zero, thus the mechanical energy is:

Since the energy is conserved, U1=U2
![\displaystyle mgH=\frac{1}{2}mv^2 \qquad\qquad [1]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20mgH%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20%20%20%20%5Cqquad%5Cqquad%20%5B1%5D)
For the speed to be double, we need to drop the snowball from a height H', and:

Operating:
![\displaystyle mgH'=4\frac{1}{2}m(v)^2 \qquad\qquad [2]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20mgH%27%3D4%5Cfrac%7B1%7D%7B2%7Dm%28v%29%5E2%20%5Cqquad%5Cqquad%20%5B2%5D)
Dividing [2] by [1]

Simplifying:

Thus:
H' = 4H
Correct choice: b 4H
it would be at either A or B.