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Step2247 [10]
4 years ago
9

How would you expect a positive particle approaching another positive particle to behave?

Chemistry
1 answer:
JulijaS [17]4 years ago
8 0

Answer:

They would produce a repulsive force to another

Explanation:

A positive particle approaching another positive particle will repulse it.

According to coulomb's law "like charges repel one another and unlike charges attract".

A charge is an intrinsic property of any matter.

When like charges e.g positive and positive or negative and negative charges are in the vicinity of one another, they repel each other.

When unlike charges; positive and negative are brought together, they simply attract one another.

Therefore, we expect that a positive particle approaching another positive particle will repel one another.  

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The fire department needs information on friction losses occurring between a water main and an open fire hydrant. At maximum mai
lisov135 [29]

Explanation:

The given data is as follows.

      P_{1} = 85 psig,   P_{2} = P_{atm} = 15 psia

        Q = 1620 gpm,  d = 2.5 inch,     l = 8 ft = 2.4384 m

According to Darey-Weisbach equation,

                        h_{l} = \frac{4fl \nu^{2}}{2gD}  ......... (1)

Value of 'f' will be decided on the basis of Reynold number.

As, it is known that R_{l} = \frac{\rho \nu d}{\mu}

where,  \mu_{water} = 10^{-3} kg/ms

As it is known that 1 gpm = \frac{1}{3.67} m^{3}/hr

So,  1 m^{3}/hr = 3.67 gpm

Therefore,   Q = 1620 \times \frac{1}{3.67}

                        = 441.4168 m^{3}/hr

                         = 0.1226 m^{3}/s

In, 1 inch = 2.54 cm = 0.0254 m

Therefore, d = 2.5 \times 0.0254 = 0.0635 m

                V = \frac{Q}{\frac{\pi}{4}d^{2}}

                    = \frac{0.1226}{0.785 \times (0.0635)^{2}}

                    = 38.73 m/s

Hence, we will calculate Reynold number as follows.

             R_{l} = \frac{1000 \times 38.73 \times 0.0635}{10^{-3}}

                             = 2459355

As R_{l} > 2000 then, it means that flow is turbulent.

As, f = 0.079 R^{-0.25}_{l}

        = 0.001994

Putting all the values into equation (1) formula as follows.    

                          h_{l} = \frac{4fl \nu^{2}}{2gD}

                                     = \frac{4 \times 0.001994 \times 2.4384 \times (38.73)^{2}}{2 \times 9.81 \times 0.0635}

                                      = 1.04069 \times 10^{5} m

Thus, we can conclude that friction loss from the main to the discharge point is 1.04069 \times 10^{5} m.

4 0
3 years ago
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