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denis23 [38]
2 years ago
15

Water (density p-1000 is discharging from through a hole at the bottom of a graduated 71 cylinder. The mass flow rate exiting th

e container may be approximated by: Where Cd is the dimensionless discharge coefficient, A, is the area of the discharge hole, g is the acceleration due to gravity, and y is the water height in the container. The diameter of the graduated cylinder is 8 cm.
For the following givens, use matlab or a spreadsheet to estimate the time to empty half the water from the container.

For partial credit, perform a simple calculation to estimate an amount of time known to be too short but greater than zero.

C,-0.6 2 ,-0.04 cm g 9.8 m/s/s

Yinitial20cm

Engineering
1 answer:
alexira [117]2 years ago
4 0

Answer:

Please see attachment

Explanation:

Please see attachment

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Air is contained in a vertical piston–cylinder assembly such that the piston is in static equilibrium. The atmosphere exerts a p
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Answer:

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b) 32 kg

Explanation:

The gauge pressure is of the gas is equal to the weight of the piston divided by its area:

p = P / A

p = m * g / (π/4 * d^2)

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p * (π/4 * d^2) = m * g

m = p * (π/4 * d^2) / g

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A particle moving on a straight line has acceleration a = 5-3t, and its velocity is 7 at time t = 2. If s(t) is the distance fro
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Given acceleration a = 5-3t, and its velocity is 7 at time t = 2, the value of s2 - s1 = 7

<h3>How to solve for the value of s2 - s1</h3>

We have

= \frac{dv}{dt} =v't = 5-3t\\\\\int\limits^a_b {v'(t)} \, dt

= \int\limits^a_b {(5-3t)} \, dt

5t - \frac{3t^2}{2} +c

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= 4+c

s(t) = \frac{5t^2}{2} -\frac{t^3}{2} +3t + c

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=(5*\frac{4}{2} -\frac{8}{2} +3*2*c)-(\frac{5}{2} *1^2-\frac{1^2}{2} +3*1*c)

= 6 + 6+c - 2+3+c

12+c-5+c = 0

7 = c

Read more on acceleration here: brainly.com/question/605631

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