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olchik [2.2K]
3 years ago
8

An irrigation canal has a rectangular cross section. At one point where the canal is 18.2 m wide and the water is 3.55 m deep, t

he water flows at 2.55 cm/s . At a second point downstream, but on the same level, the canal is 16.3 m wide, but the water flows at 11.6 cm/s . How deep is the water at this point
Physics
1 answer:
Luden [163]3 years ago
3 0

Answer:

Explanation:

Rate of volume flow at two points will be same at two points .

A₁ V₁ = A₂V₂

A₁ and A₂ are area of cross section at two points and V₁ and V₂ are velocities .

A₁ = 18.2 x 3.55 = 64.61 m²

V₁ = 2 .55 x 10⁻² m/s

A₂ = 16.3 x d = 16.3 d  m²

d is depth at second point .

V₂ = 11.6 x 10⁻² m/s

64.61 m² x 2 .55 x 10⁻² m/s = 16.3 d  m² x 11.6 x 10⁻² m/s

d = .87 m

so canal is .87 m deep.

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allsm [11]

Answer:

a) α = -65,2 rad/s².

b) t = 2,57 s.

Explanation:

a) La aceleración angular se puede calcular usando la siguiente ecuación:

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En donde:

\omega_{f}: es la velocidad angular final =  2000 rpm = 209,4 rad/s

\omega_{0}: es la velocidad angular inicial = 3600 rpm = 377,0 rad/s

α: es la aceleración angular=?

θ: es el desplazamiento o número de vueltas = 120 rev = 754,0 rad

Las conversiones de unidades se hicieron sabiendo que 1 revolución = 2π radianes y que 1 minuto = 60 segundos.  

Resolviendo la ecuación (1) para α, tenemos:

\alpha = \frac{\omega_{f}^{2} - \omega_{0}^{2}}{2\theta} = \frac{(209,4 rad/s)^{2} - (377,0 rad/s)^{2}}{2*754,0 rad} = -65,2 rad/s^{2}  

Entonces, la aceleración angular es -65,2 rad/s². El signo negativo se debe a que el motor está desacelerando.  

b) El tiempo transcurrido se puede encontrar como sigue:

\omega_{f} = \omega_{0} + \alpha t

Resolviendo para t, tenemos:

t = \frac{\omega_{f} - \omega_{0}}{\alpha} = \frac{209,4 rad/s - 377,0 rad/s}{-65,3 rad/s^{2}} = 2,57 s

Por lo tanto, el tiempo transcurrido fue 2,57 s.

Espero que te sea de utilidad!

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