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3 years ago

List and describe three basic Boolean operations commonly used in

Engineering

1 answer:

valentinak56 [21]3 years ago

4 0

Answer:

There are three basic Boolean operations: JOIN (Union), CUT (Difference), and INTERSECT. ... The CUT operation subtracts the volume of one solid object from the other solid object. The INTERSECT operation keeps only the volume common to both solid objects.

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A. Calculate the fraction of atom sites that are vacant for lead at its melting temperature of 327°C (600 K). Assume an energy f

NISA [10]

Answer:

a. Fraction of Atom = 2.41E-5 when T = 600K

b. Fraction of Atom = 5.03E-10 when T = 298K

Explanation:

Given

T = Temperature = 600K

Qv = Energy for formation = 0.55eV/atom

To calculate the fraction of atom sites, we make use of the following formula

Nv/N = exp(-Qv/kT)

Where k = Boltzmann Constant = 8.62E-5eV/K

Nv/N = exp(-0.55/(8.62E-5 * 600))

Nv/N = 0.000024078672493307

Nv/N = 2.41E-5

b. When T = 298K

Nv/N = exp(-0.55/(8.62E-5 * 298))

Nv/N = 5.026591237904E−10

Nv/N = 5.03E-10 ----- Approximated

6 0

4 years ago

Consider a 1.80-m-tall man standing vertically in water and completely submerged in a pool. Determine the difference between the

defon

Answer:

17.658 kPa

Explanation:

The hydrostatic pressure of a fluid is the weight of a column of that fluid divided by the base of that column.

$P = \frac{weight}{base}$

Also, the weight of a column is its volume multiplied by it's density and the acceleration of gravity:

$weight = \delta * v * g$

Meanwhile, the volume of a column is the area of the base multiplied by the height:

$V = base * h$

Replacing:

$P = \frac{\delta * base * h * g}{base}$

The base cancels out, so:

$P = \delta * h * g$

The pressure depends only on the height of the fluid column, the density of the fluid and the gravity.

If you have two point at different heights (or depths in the case of objects submerged in water) each point will have its own column of fluid exerting pressure on it. Since the density of the fluid and the acceleration of gravity are the same for both points (in the case of hydrostatics density is about constant for all points, it is not the case in the atmosphere), we can write:

$\Delta P = \rho * g * (h1 - h2)$

We do not know at what depth the man of this problem is, but it doesn't matter, because we know the difference in height of the two points of interes (h1 - h2) = 1.8 m. So:

$\Delta P = 9.81 \frac{m}{s^{2} } * 1000 \frac{kg}{m^3} * 1.8 m = 17658 Pa = 17.658 kPa$