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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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The marble lands at a distance of 1.36 m from the base of the table.

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The motion of the marble falling off the table is a projectile with initial velocity in the horizontal direction only.

The motion can be solved in two directions, the horizontal and vertical direction.

Along the vertical direction, the initial velocity is 0 m/s as it has only horizontal component initially. Also acceleration in the vertical direction is acceleration due to gravity.

Let us use equation of motion in vertical direction.

y-y_0=v_{0y}t+\frac{1}{2}at^2

Where,

v_{0y}\rightarrow \textrm{vertical component of the initial velocity}.\\y\rightarrow \textrm{final position of the marble}\\y_0\rightarrow \textrm{initial vertical position}\\a_y\rightarrow \textrm{acceleration in the vertical direction}\\t\rightarrow \textrm{time taken to reach bottom}

Now, plug in 0 for y, 1.25 for y_0, 0 for v_{0y}, -9.8 for a_y. This gives,

0-1.25=0+\frac{1}{2}(-9.8)t^2\\-1.25=-4.9t^2\\t^2=\frac{-1.25}{-4.9}\\t^2=0.255\\t=\sqrt{0.255}=0.501\ s

Therefore, time to reach bottom is 0.501 s.

Now, consider the horizontal motion. There is no acceleration in the horizontal direction. So, distance is given as the product of horizontal velocity and time taken.

Horizontal distance covered by the marble is given as:

x=v_{0x}\times t=2.71\times 0.501=1.36 \textrm{ m}

Therefore, the marble lands at a distance of 1.36 m from the base of the table.

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3 years ago
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