Answer: 71.7 KJ
Explanation:
The rotational kinetic energy of a rotating body can be written as follows:
Krot = ½ I ω2
Now, any point on the rim of the flywheel, is acted by a centripetal force, according to Newton’s 2nd Law, as follows:
Fc = m. ac
It can be showed that the centripetal acceleration, is related with the angular velocity and the radius, as follows:
ac = ω2 r
We know that this acceleration has a limit value, so , we can take this limit to obtain a maximum value for the angular velocity also.
As the flywheel is a solid disk, the rotational inertia I is just ½ m r2.
Replacing in the expression for the Krot, we have:
Krot= ½ (1/2 mr2.ac/r) = ¼ mr ac = ¼ 67.0 Kg. 1.22 m . 3,510 m/s2 = 71. 7 KJ
It takes a noun and a verb to make a complete sentence.
There isn't a single verb in a), b), or c).
"Affords" is the verb (predicate) in d)., the only complete sentence.
Answer:
The answer to your question is below
Explanation:
If Oxygen has an atomic number of 8, we can conclude that:
- If neutral, it has 8 protons and 8 electrons
- It has 8 neutrons and its atomic mass is 16 (A = 8 + 8)
- It must be located in the group VI A
- Its valence number must be 6
- Its Oxidation number is -2
Answer:
El mango llega al suelo a una velocidad de 329.982 metros por segundo.
Explanation:
El mango experimenta un movimiento de caída libre, es decir, un movimiento uniformemente acelerado debido a la gravedad terrestre, despreciando los efectos de la viscosidad del aire y la rotación planetaria. Entonces, la velocidad final del mango, es decir, la velocidad con la que llega al suelo, se puede determinar mediante la siguiente fórmula cinemática:
(1)
Donde:
- Velocidad inicial, en metros por segundo.
- Velocidad final, en metros por segundo.
- Aceleración gravitacional, en metros por segundo al cuadrado.
- Tiempo, en segundos.
Si sabemos que 
, 
y 
, entonces la velocidad final del mango es:
El mango llega al suelo a una velocidad de 329.982 metros por segundo.
