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rodikova [14]
2 years ago
10

6

Engineering
2 answers:
grin007 [14]2 years ago
8 0

Answer:

(1)speed at which ball B was thrown upward-13.95m/s (2) velocity pf a=14m/s and b=7.25 m/s

Explanation:

Eduardwww [97]2 years ago
7 0

Answer:

ball a is released from rest at a height of 20 m

Explanation:

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Suppose the working pressure for a boiler is 10 psig, then what is the corresponding absolute pressure?
yanalaym [24]

Answer:

The corresponding absolute pressure of the boiler is 24.696 pounds per square inch.

Explanation:

From Fluid Mechanics, we remember that absolute pressure (p_{abs}), measured in pounds per square inch, is the sum of the atmospheric pressure and the working pressure (gauge pressure). That is:

p_{abs} = p_{atm}+p_{g} (1)

Where:

p_{atm} - Atmospheric pressure, measured in pounds per square inch.

p_{g} - Working pressured of the boiler (gauge pressure), measured in pounds per square inch.

If we suppose that p_{atm} = 14.696\,psi and p_{g} = 10\,psi, then the absolute pressure is:

p_{abs} = 14.696\,psi+10\,psi

p_{abs} = 24.696\,psi

The corresponding absolute pressure of the boiler is 24.696 pounds per square inch.

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3 years ago
A gas stream contains 4.0 mol % NH3 and its ammonia content is reduced to 0.5 mol % in a packed absorption tower at 293 K and 10
bagirrra123 [75]

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Explanation:

Step by step solved solution is given in the attached document.

8 0
3 years ago
What is an example of a class 2 lever?
9966 [12]

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A wheelbarrow, a bottle opener, and an oar are examples of second class levers

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2 years ago
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g A steel water pipe has an inner diameter of 12 in. and a wall thickness of 0.25 in. Determine the longitudinal and hoop stress
zvonat [6]

Answer:

a) \mathbf{\sigma _ 1 = 4800 psi}

     \mathbf{ \sigma _2 = 0}

b)\mathbf{\sigma _ 1 = 6000 psi}

  \mathbf{ \sigma _2 = 3000 psi}

Explanation:

Given that:

diameter d = 12 in

thickness t = 0.25 in

the radius = d/2 = 12 / 2 = 6 in

r/t = 6/0.25 = 24

24 > 10

Using the  thin wall cylinder formula;

The valve A is opened and the flowing water has a pressure P of 200 psi.

So;

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = 0

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{200(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 4800 psi}

b)The valve A is closed and the water pressure P is 250 psi.

where P = 250 psi

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = \frac{Pd}{4t}

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{250*(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 6000 psi}

\sigma _2 = \frac{Pd}{4t} \\ \\  \sigma _2 = \frac{250(12)}{4(0.25)}

\mathbf{ \sigma _2 = 3000 psi}

The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below

8 0
3 years ago
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