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lilavasa [31]
3 years ago
15

Motor oil , with a viscosity of 0 . 250 Ns / m2 , is flowing through a tube that has a radius of 5 . 00 mm and is 25 . 0 cm long

. The drop in pressure is 300 kPa . What is the volume of oil flowing through the tube per unit time ?
Physics
1 answer:
Thepotemich [5.8K]3 years ago
3 0

Answer:

1.1775 x 10^-3 m^3 /s

Explanation:

viscosity, η = 0.250 Ns/m^2

radius, r = 5 mm = 5 x 10^-3 m

length, l = 25 cm = 0.25 m

Pressure, P = 300 kPa = 300000 Pa

According to the Poisuellie's formula

Volume flow per unit time is

V=\frac{\pi \times Pr^{4}}{8\eta l}

V=\frac{3.14 \times 300000\times \left ( 5\times 10^{-3} \right )^{4}}{8\times 0.250\times 0.25}

V = 1.1775 x 10^-3 m^3 /s

Thus, the volume of oil flowing per second is 1.1775 x 10^-3 m^3 /s.

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A quarterback passes a football from height h = 2.1 m above the field, with initial velocity v0 = 13.5 m/s at an angle θ = 32° a
SOVA2 [1]

Answer:

a)    x = v₀² sin 2θ / g

b)    t_total = 2 v₀ sin θ / g

c)    x = 16.7 m

Explanation:

This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

        sin θ = v_{oy} / vo

        cos θ = v₀ₓ / vo

         v_{oy} = v_{o} sin θ

         v₀ₓ = v₀ cos θ

         v_{oy} = 13.5 sin 32 = 7.15 m / s

         v₀ₓ = 13.5 cos 32 = 11.45 m / s

a) In the x axis there is no acceleration so the velocity is constant

         v₀ₓ = x / t

          x = v₀ₓ t

the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero

          v_{y} = v_{oy} - gt

          0 = v₀ sin θ - gt

          t = v_{o} sin θ / g

         

we substitute

       x = v₀ cos θ (2 v_{o} sin θ / g)

       x = v₀² /g      2 cos θ sin θ

       x = v₀² sin 2θ / g

at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,

b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time

at the highest point the vertical speed is zero

          v_{y} = v_{oy} - gt

          v_{y} = 0

           t = v_{oy} / g

           t = v₀ sin θ / g

as the time to get on and off is the same the total time or flight time is

           t_total = 2 t

           t_total = 2 v₀ sin θ / g

c) we calculate

          x = 13.5 2 sin (2 32) / 9.8

          x = 16.7 m

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3 years ago
3) Ęplain why muddy water is<br> Heterogeneous<br> mixture
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Answer:

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