(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
#SPJ1
Explanation:
Track and Field is a sport, which is includes disciplines of running, jumping, and throwing events. The sport traces back to Ancient Greece. The first recorded examples of this sport were at the Ancient Greek Olympics. In Ancient Greece, only one event was contested, the stadion footrace. Later on, the game expanded to more events.Events of track and field are divided into three: track events, field events, and combined events. Track events consist of Sprints, middle-distance, long distance, hurdles and relays; Field events consist of jumps and throws; while combined events consist of pentathlon, heptathlon, and decathlon. Track and field is usually played outdoors in stadiums. The usual features of a track and field stadium are the outer running track, and the field within the track
Answer:
zero
rank the magnitude of the average velocity over the first 2 second
Answer:
a) The current is i = 1.2 A
b) The charge is Q = 17280 C
c) The energy is E = 43200 J
Explanation:
a) The current is given by the ohm's law wich is:
i = V/R = 3/2.5 = 1.2 A
b) Since the charge is steady we can use the following equation to find the charge amount in that time:
i = Q/t
Q = t*i
Where t is in seconds, so we have 4h * 3600 = 14400 s
Q = 1.2*14400 = 17280 C
c) The energy is the power delivered to the toy multiplied by the time:
P = 1.2*2.5 = 3 W
E = P*t = 3*14400 = 43200 J
