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1 year ago

5

Glycerin at a temperature of 30 degrees celcius flows at a rate of 8×10−6m3/sthrougha horizontal tube with a 30mmdiameter. what

is the pressure drop in pascals per100m?

1 answer:

marusya05 [52]1 year ago

3 0

The pressure drop in pascal is 3.824*10^4 Pascals.

To find the answer, we need to know about the Poiseuille's formula.

<h3>How to find the pressure drop in pascal?</h3>

We have the Poiseuille's formula,

$Q=\frac{\pi r^4P}{8\beta l}$

where, Q is the rate of flow, P is the pressure drop, r is the radius of the pipe, is the coefficient of viscosity (0.95Pas-s for Glycerin) and l being the length of the tube.

By substituting values and rearranging we will get the pressure drop as,

$P=3.284Pascals$

Thus, we can conclude that, the pressure drop in pascal is 3.824*10^4.

Learn more about the Poiseuille's formula here:

brainly.com/question/13180459

#SPJ4

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Answer:

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To find the mas of the box $m_2$ .

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Momentum before collision = Momentum after collision

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$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$

We can plugin the given value to find $m_2$

$(0.0427\times 22.3)+(m_2\times 0)=(0.0427\times (-11.5))(m_2\times 1.53)$

$0.9522+0=-0.4911+1.53m_2$

Adding both sides by 0.4911

$0.9522+0.4911=-0.4911+0.4911+1.53m_2$

$1.4433=1.53m_2$

Dividing both sides by 1.53.

$\frac{1.4433}{1.53}=\frac{1.53m_2}{1.53}$

$0.9433=m_2$

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3 years ago

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Given:

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- Potential difference p.d = 10 V

Find:

What speed does particle #2 attain?

Solution:

- The force on a charged particle in an electric field is given by:

F = Q*V / r

Where, r is the distance from one end to another.

- The Net force acting on a charge accelerates it according to the Newton's second equation of motion:

F_net = m*a

- Equate the two expressions:

a = Q*V / m*r

- The speed of the particle in an electric field is given by third kinetic equation of motion.

v_f^2 - v_i^2 = 2*a*r

Where, v_f is the final velocity,

v_i is the initial velocity = 0

v_f^2 - 0 = 2*a*r

Substitute the expression for acceleration in equation of motion:

v_f^2 = 2*(Q*V / m*r)*r

v_f^2 = 2*Q*V / m

v_f = sqrt (2*Q*V / m)

- The velocity of first particle is v:

v = sqrt (20*Q / m)

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