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docker41 [41]
3 years ago
9

A pump uses a piston of 15 cm diameter that moves at 2.0 cm/s as it pushes a fluid through a pipe. what is the speed of the flui

d when it enters a portion of the pipe that is 3.0 mm in diameter? treat the fluid as ideal and incompressible.
Physics
1 answer:
Nitella [24]3 years ago
4 0
The important point here is that volumetric flow rate in the pump and the pipe is the same.

Q = AV, where Q = Volumetric flow  rate, A = Cross sectional area, V = velocity

Q (pump) = (π*15^2)/4*2 = 353.43 cm^3/s
Q (pipe) = (π*(3/10)^2)/4*V = 0.071V

Q (pump) = Q (pipe)
0.071V = 353.43 => V = 5000 cm/s

Therefore, the flow  of water in the pipe is 5000 cm/s.
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A. The applied force should be the same size as the friction force

Explanation:

Whenever we apply a force to an object it moves if the force applied to that object is unbalanced and there is no force or a lesser force to counter it. According to Newton's Second Law of motion, when an unbalanced force is applied to an object it produces an acceleration in the object in its own direction. So, the two forces acting on this box are the frictional force and the applied force in horizontal direction. In order to move the box at constant speed, the applied force must first, overcome the frictional force, so the object can start its motion. Since, the motion has constant velocity, it means no acceleration. So, the force must be balanced in order to avoid acceleration as a consequence of Newton's Second Law of motion. Therefore, the correction in this case will be:

<u>A. The applied force should be the same size as the friction force</u>

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A 12. 0 cm object is 9. 0 cm from a convex mirror that has a focal length of -4. 5 cm. What is the distance of the image from th
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A mirror equation is an equation that relates object distance and image distance to focal length. The distance of the image from the mirror will be -3 cm.

<h3>What is a mirror equation?</h3>

A mirror equation is an equation that relates object distance and image distance to focal length. It's sometimes referred to as a mirror formula.

\rm \frac{1}{v} +\frac{1}{u} =\frac{1}{f} \\\\

The given data in the problem is;

f is the focal length =  -4. 5 cm

u is the object distance=12. 0 cm

v is the image distance =?

\rm \frac{1}{v} +\frac{1}{12.} =\frac{1}{-4.5} \\\\ \rm \frac{1}{v}= \frac{1}{-3} \\\\ \rm v= -3cm

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To learn more about the mirror equation refer to the link;

https://brainly.in/question/31545592

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