(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:
Explanation:
Both these questions are based on the Universal Law of Gravitation, which is given by :
F = Gm1m2 / r²
2) F = 6.67 x 10⁻¹¹ x 8 x 10³ x 1.5 x 10³ / 1.5 x 1.5
F = 6.67 x 10⁻⁵ x 8 / 1.5
F = 35.57 x 10⁻⁵ N
3) F = 6.67 x 10⁻¹¹ x 7.5 x 10⁵ x 9.2 x 10⁷ / 7.29 x 10⁴
F = 6.67 x 10⁻³ x 7.5 x 9.2 / 7.29
F = 63.13 x 10⁻³ N
Answer:
6
Explanation:
because if 3kg is 8m/s then 16 divided by 2 is 8 means add 3 to 3 and get the final answer
Answer:
when a element of 1 group take part in reaction, its atom looses outer electron and form positively charged ions called Cation.
Explanation: