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Damm [24]
3 years ago
11

While traveling on a dirt road, the bottom of a car hits a sharp rock anda small hole develops at the bottom of its gas tank. If

the height of the gasoline in thetank is 35 cm, determine the initial velocity of the gasoline at the hole. Discuss howthe velocity will change with time and how the flow will be affected it the lid of thetank is closed tightly.
Physics
1 answer:
tester [92]3 years ago
5 0

Answer:

6.86 m/s

Explanation:

velocity of the gasoline through the hole would be

v = \sqrt {2gh}\\v = \sqrt {2\times 9.8\times 0.35} =6.86 m/s

As the height would decrease, the velocity of the flow would decrease.

If the lid of the tank would be closed, the pressure would reduce. so, again the velocity of the flow would reduce.

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Un engrane que gira con una velocidad de 20 rad/s, es acelerado durante 5 segundos hasta alcanzar una velocidad de 35 rad/s
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Answer:

a) La aceleración angular es: \alpha=2\: rad/s^{2}

b) El engranaje gira 125 radianes.

c) El engranaje hara aproximadamente 20 revoluciones.

Explanation:

a)

La aceleración angular se define como:

\alpha=\frac{\Delta \omega}{\Delta t}

Donde:

  • Δω es la diferencia de velocidad angular (en otras palabras ω(final)-ω(inicial))
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\alpha=\frac{35-25}{5}

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b)

El desplazamiento angular puede ser calculado usando la siguiente ecuación:

\theta=\theta_{i}+\omega_{i}t+\frac{1}{2}\alpha t^{2}

Aqui el angulo inicial es 0, por lo tanto.

\theta=20(5)+\frac{1}{2}(2)(5)^{2}

\theta=125\: rad

El engranaje gira 125 radianes.

c)

Lo que debemos hacer aquí es convertir radianes a revoluciones.

Recordemos que 2π rad = 1 rev

Entonces:

\theta=125\: rad \times \frac{1\: rev}{2\pi\: rad}=19.89\: rev

Por lo tanto el engranaje hara aproximadamente 20 revoluciones.

Espero te haya sido de ayuda!

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