Answer:
a) La aceleración angular es: 
b) El engranaje gira 125 radianes.
c) El engranaje hara aproximadamente 20 revoluciones.
Explanation:
a)
La aceleración angular se define como:
Donde:
- Δω es la diferencia de velocidad angular (en otras palabras ω(final)-ω(inicial))
- Δt es el tiempo en el que occure el cambio de velocidad angular
b)
El desplazamiento angular puede ser calculado usando la siguiente ecuación:
Aqui el angulo inicial es 0, por lo tanto.
El engranaje gira 125 radianes.
c)
Lo que debemos hacer aquí es convertir radianes a revoluciones.
Recordemos que 2π rad = 1 rev
Entonces:
Por lo tanto el engranaje hara aproximadamente 20 revoluciones.
Espero te haya sido de ayuda!
The velocity of the body is zero; option A
<h3>What is the motion of an oscillating body?</h3>
The motion of an oscillating body is known as simple harmonic motion.
Simple harmonic motion involves a periodical motion of a body whose acceleration is directed towards a fixed point.
For a body that is oscillating up and down at the end of a spring, considering when the body is at the top of its up-and-down motion, the velocity of the body at the top and down is zero since the body comes to rest at the top and down position of its motion.
In conclusion, oscillating bodies undergo simple harmonic motion.
Learn more about simple harmonic motion at: brainly.com/question/24646514
#SPJ1
I am going to need a picture for this question
Answer:
The coefficient of kinetic friction between the puck and the ice is 0.11
Explanation:
Given;
initial speed, u = 9.3 m/s
sliding distance, S = 42 m
From equation of motion we determine the acceleration;
v² = u² + 2as
0 = (9.3)² + (2x42)a
- 84a = 86.49
a = -86.49/84
|a| = 1.0296
= ma
where;
Fk is the frictional force
μk is the coefficient of kinetic friction
N is the normal reaction = mg
μkmg = ma
μkg = a
μk = a/g
where;
g is the gravitational constant = 9.8 m/s²
μk = a/g
μk = 1.0296/9.8
μk = 0.11
Therefore, the coefficient of kinetic friction between the puck and the ice is 0.11
