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Sholpan [36]
3 years ago
9

Based on the table, what is the advantage for using 16-inch spacing compared to 24-

Physics
1 answer:
Rzqust [24]3 years ago
6 0

The advantage for using 16-inch spacing is that ; ( B ) At a given load, the 16-inch spacing can support a larger span

The spacing along a span in construction engineering is determined by the amount and mass of load the span will either carry or support during and after construction.

A smaller span spacing can support a given load more efficiently than a larger span spacing because when the spacing is smaller it can support a larger span better than a larger span spacing.

Hence we can conclude that the advantage for using 16-inch spacing is that At a given load, the 16-inch spacing can support a larger span.

learn more : brainly.com/question/22020642

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when an object slides over a smooth horizontal surface, how does the force of friction depend on the surface area of blocks that
Marina CMI [18]

Answer: with a greater surface area, there will be a greater force of friction

Explanation:

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3 years ago
How does temperature change as you go from the surface toward the center of the earth?
Nastasia [14]
Because lava is in the center and lavas hot
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4 years ago
Write one advantage of MKS system over CGS system.​
Temka [501]
More convenient
More commonly used
7 0
3 years ago
Minute after minute, hour after hour, day after day, ocean waves continue to splash onto the shore. Explain why the beach is not
MAVERICK [17]

Answer: C. Ocean waves can only bring energy to the shore; the particles of the medium (water) simply oscillate about their fixed position.

Explanation:

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It should be noted that even a single drop of water cannot be brought by the ocean wave to the shore from the middle of the ocean.

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4 0
3 years ago
Calculate the energy (in eV/atom) for vacancy formation in some metal, M, given that the equilibrium number of vacancies at 296o
Schach [20]

Explanation:

The given data is as follows.

       Temperature of metal = 296^{o}C = (296 + 273) K

                                            = 569 K

     Density of the metal = 8.85 g/cm^{3} = 8.85 \times 10^{-6} g/m^{3}      (as 1 cm^{3} = 10^{-6} m^{3})

     Atomic mass = 51.40 g/mol

    Vacancies = 9.19 \times 10^{23} m^{-3}

Formula to calculate the number of atomic sites is as follows.

           n = \frac{\rho \times N_{A}}{\text{atomic weight}}

              = \frac{8.85 \times 10^{-6} \times 6.022 \times 10^{23}}{51.40 g/mol}

              = 1.036 \times 10^{17} atom/m^{3}

Now, we will calculate the energy as follows.

                E = -KT \times ln (\frac{\text{no. of vacancies}}{\text{no. of atomic sites}})

where,    K = 8.62 \times 10^{-5}

         E = -8.62 \times 10^{-5} \times 569 K \times ln (\frac{9.19 \times 10^{23}}{1.036 \times 10^{17} atom/m^{3}})

               = 78.46 eV/atom

Therefore, we can conclude that energy (in eV/atom) for vacancy formation in given metal, M, is 78.46 eV/atom.

6 0
4 years ago
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