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!
Answer:
a)
b)
Explanation:
From the exercise we know initial velocity, initial height
º
a) The range of the stone is defined by how far does it goes. From the theory of <u><em>free falling objects</em></u>, we have:
The stone strike the water at y=0
Solving for t, using the quadratic formula
or
Since time can't be negative, the answer is t=3.06s
Now, we can calculate the <u>range of the stone</u>
b) We can calculate the <u>velocity</u> were the stone strike the water using the following formula
The negative sign indicates that the stone is going down
Answer:
Required horizontal distance is 0.9279 meters
Explanation:
The situation is represented in the attached figure
The horizontal distance can be seen to be equal to
In the upper triangle we have
Now the angle can be calculated using Snell's law
By snell's Law we have
Since light comes from air thus
Light enter's the water thus we have
Applying values we calculate as
Now in the attached figure we have
Solving for we get
thus the required horizontal distance is
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If we are talking on the force being exerted by a segment of a rope of lenght R on the right on a point M which is being also pulled from the Left by a segment of rope R as shown in the figure attached. Then we invoke Newton's Third Law:
"Any force exerted by an object (in this case a segment of the rope) also suffers a equal and opposite force".
If we pick
whis is the tension exerted by the right segment then the left segment will also exert an equal and opposite force so we have that