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Serggg [28]
1 year ago
5

How much water will flow in 30 secs through 200 mm of capillary tube of 1.50 mm in diameter, if the pressure difference across t

he tube is 6660N/m² and viscosity is 8.01x10 kg/ms​
Physics
1 answer:
Paladinen [302]1 year ago
5 0

The water outflow in 30 secs through 200 mm of the capillary tube is mathematically given as

Qo=1.6 \times 10^{2} \mathrm{~mL}

<h3>What is the water outflow in 30 secs through 200 mm of the capillary tube?</h3>

\begin{aligned}\Delta P &=6660 \mathrm{~m} / \mathrm{m}^{2} \\\mu &=8.01 \times 10^{-4} \text { Pas } \\t &=30 \mathrm{~s} \\L &=200 \mathrm{~mm}=200 \times 10^{-3} \mathrm{~m} \\D &=1.5 \mathrm{~mm}=1.5 \times 10^{-3} \mathrm{~m} \Rightarrow \gamma=\frac{1.5 \times 10^{-3}}{2} \mathrm{~m}\end{aligned}

Generally, the equation for Rate of flow of Liquid is  mathematically given as

\\$$Q=\frac{\pi r^{4} \times \Delta P}{8 \mu L}

$$

Where dP is pressure difference r is the radius

\mu is the viscosity of water

L is the length of the pipe

Q=\frac{\pi \times\left(\frac{1.5 \times 10^{-3}}{2}\right)^{4} \times 6660}{8 \times 8.01 \times 10^{-4} \times 200 \times 10^{-3}}

Q=5.2 \mathrm{~mL} / \mathrm{s}

In $30s the quantity that flows out of the tube

&Qo=5.2 \times 30 \\&Qo=1.6 \times 10^{2} \mathrm{~mL}

In conclusion, the quantity that flows out of the tube

Qo=1.6 \times 10^{2} \mathrm{~mL}

Read more about the flows rate

brainly.com/question/27880305

#SPJ1

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3 years ago
Two particles P and Q each moves towards ther along Straight Line (M)(N), 51m long. P starts from M with velocity 5m/s and const
bagirrra123 [75]

The time required by the particles are as follows:

a. t = 1.5 seconds

b. t = 3 seconds

c. t = 0.4 seconds

<h3>What is the time required?</h3>

The time required for the particles to be at several distances apart is calculated using the equation of motion given below:

S = ut + \frac{1}{2}at^{2}

a) Time required to be 30 m apart:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51 - 30

S1 + S2 = 21

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 21

2t^2 + 11t - 21 = 0

Solving for time, t by factorization, t = 1.5 seconds

b) Time required to meet:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 51

2t^2 + 11t - 51 = 0

Solving for time, t by factorization, t = 3 seconds

c) Time required for velocity of P is ¾ of the velocity of Q:

Using the equation of motion: V = u + at

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Substituting the values:

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5t = 2

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