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Zigmanuir [339]
3 years ago
10

Match the words with their definitions

Engineering
1 answer:
Pavlova-9 [17]3 years ago
5 0

Wooden 2 X 8's, 10's, or 12's that run parallel to one another and support a floor or ceiling, and supported in turn by larger beams, girders, or bearing walls- <u>Joist</u>

Explanation:

  1. Wooden 2 X 8's, 10's, or 12's that run parallel to one another and support a floor or ceiling, and supported in turn by larger beams, girders, or bearing walls- <u>Joist</u>
  2. A point where a bearing/structural weight is concentrated and transferred to the foundation-<u>Point Load</u>
  3. Includes all exterior walls and any interior wall that is aligned above a support beam or girder-<u>Load Bearing Wall</u>
  4. Plywood substitute made of coarse sawdust that is mixed with resin and pressed into sheets-<u>Particleboard</u>
  5. The replacement of excavated earth into a trench around or against a basement/crawl space foundation wall- <u>Backfill</u>
  6. Continuous 8m" or 10" thick concrete pad installed before and supports the foundation wall or monopost-<u>Footing</u>
  7. Ribbed steel bars installed in foundation concrete walls, footers and poured in place concrete structures designed to strengthen concrete-<u>Rebar</u>
  8. The end, upper, triangular area of a home, beneath the roof-<u>Gable</u>
  9. A manufactured 4X8 wood panel made out of 1-2 wood chips and glue. Often used as a substitute for plywood-<u>OSB</u>
  10. The incline slope of a roof or the ratio of the total width of a house-<u>Pitch</u>
  11. A manufactured structual wood beam. It is constructed of pressure and adhesive wood strands of wood-<u>Microlam</u>

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About what thickness of aluminum is needed to stop a beam of (a) 2.5-MeV electrons, (b) 2.5-MeV protons, and (c) 10-MeV alpha pa
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The thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.

<h3>Thickness of the aluminum</h3>

The thickness of the aluminum can be determined using from distance of closest approach of the particle.

K.E = \frac{2KZe^2}{r}

where;

  • Z is the atomic number of aluminium  = 13
  • e is charge
  • r is distance of closest approach = thickness of aluminium
  • k is Coulomb's constant = 9 x 10⁹ Nm²/C²
<h3>For 2.5 MeV electrons</h3>

r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (1.6\times 10^{-19})^2}{2.5 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m

<h3>For 2.5 MeV protons</h3>

Since the magnitude of charge of electron and proton is the same, at equal kinetic energy, the thickness will be same. r = 1.5 x 10⁻¹⁴ m.

<h3>For 10 MeV alpha-particles</h3>

Charge of alpah particle = 2e

r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (2 \times 1.6\times 10^{-19})^2}{10 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m

Thus, the thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.

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