(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:
There are several modalities, or methods, of treatment: individual therapy, group therapy, couples therapy, and family therapy
Explanation:
Answer:
The weight of the girl = 1045.86 kg/m³
Explanation:
Density: This can be defined as the ratio of the mass of a body to the volume of that body. The S.I unit of density is kg/m³.
From Archimedes principle,
R.d = Density of the person/Density of water = Weight of the person in air/Upthrust.
⇒ D₁/D₂ = W/U............................... Equation 1.
Where D₁ = Density of the person, D₂ = Density of water, W = Weight of the person in air, U = Upthrust in water.
Making D₁ the subject of the equation,
D₁ = D₂(W/U)................................... Equation 2
<em>Given: D₂ = 1000 kg/m³ , W = 509.45 N, U = lost in weight = weight in air - weight in water = 509.45 - 22.34 = 487.11 N</em>
<em>Substituting these values into equation 2</em>
D₁ = 1000(509.45/487.11)
D₁ = 1045.86 kg/m³
Thus the weight of the girl = 1045.86 kg/m³
<em></em>
Answer:
m = 3 kg
The mass m is 3 kg
Explanation:
From the equations of motion;
s = 0.5(u+v)t
Making t thr subject of formula;
t = 2s/(u+v)
t = time taken
s = distance travelled during deceleration = 62.5 m
u = initial speed = 25 m/s
v = final velocity = 0
Substituting the given values;
t = (2×62.5)/(25+0)
t = 5
Since, t = 5 the acceleration during this period is;
acceleration a = ∆v/t = (v-u)/t
a = (25)/5
a = 5 m/s^2
Force F = mass × acceleration
F = ma
Making m the subject of formula;
m = F/a
net force F = 15.0N
Substituting the values
m = 15/5
m = 3 kg
The mass m is 3 kg