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3 years ago

What are the important things to remember when arriving for an interview?

Engineering

1 answer:

ludmilkaskok [199]3 years ago

8 0

Answer: That you are dressed appropriately, to speak in a formal manner, and to be confident in your answers.

You might be interested in

Consider the diffusion of water vapor through a polypropylene (PP) sheet 1 mm thick. The pressures of H2O at the two faces are 3

Neko [114]

Answer:

$\boxed{0.000000266 \frac {cm^{3}.STP}{cm^{2}.s}}$

Explanation:

Diffusion flux of a gas, J is given by

$J=P_m\frac {\triangle P}{\triangle x}$ where $P_m$ is permeability coefficient, $\triangle$ P is pressure difference and x is thickness of membrane.

The pressure difference will be 10,000 Pa- 3000 Pa= 7000 Pa

At 298 K, the permeability coefficient of water vapour through polypropylene sheet is $38\times 10^{-13}(cm^{3}. STP)(cm)/(cm^{2}.s.Pa)$

Since the thickness of sheet is given as 1mm= 0.1 cm then

$J=38\times 10^{-13}(cm^{3}. STP)(cm)/(cm^{2}.s.Pa)\times \frac {7000 pa}{0.1cm}=0.000000266 \frac {cm^{3}.STP}{cm^{2}.s}$

Therefore, the diffusion flux is $\boxed{0.000000266 \frac {cm^{3}.STP}{cm^{2}.s}}$

7 0

3 years ago

The largest class of errors in software engineering can be attributed to _______.

Ilya [14]

Answer:

improper imput validation

Explanation:

6 0

2 years ago

In 1945, the United States tested the world’s first atomic bomb in what was called the Trinity test. Following the test, images

Zarrin [17]

Answer:

$r=K A^{1/5} \rho^{-1/5} t^{2/5}$

$A= \frac{r^5 \rho}{t^2}$

$A=1.033x10^{21} ergs *\frac{Kg TNT}{4x10^{10} erg}=2.58x10^{10} Kg TNT$

Explanation:

Notation

In order to do the dimensional analysis we need to take in count that we need to conditions:

a) The energy A is released in a small place

b) The shock follows a spherical pattern

We can assume that the size of the explosion r is a function of the time t, and depends of A (energy), the time (t) and the density of the air is constant $\rho_{air}$ .

And now we can solve the dimensional problem. We assume that L is for the distance T for the time and M for the mass.

[r]=L with r representing the radius

[A]= $\frac{ML^2}{T^2}$ A represent the energy and is defined as the mass times the velocity square, and the velocity is defined as $\frac{L}{T}$

[t]=T represent the time

$[\rho]=\frac{M}{L^3}$ represent the density.

Solution to the problem

And if we analyze the function for r we got this:

$[r]=L=[A]^x [\rho]^y [t]^z$

And if we replpace the formulas for each on we got:

$[r]=L =(\frac{ML^2}{T^2})^x (\frac{M}{L^3})^y (T)^z$

And using algebra properties we can express this like that:

$[r]=L=M^{x+y} L^{2x-3y} T^{-2x+z}$

And on this case we can use the exponents to solve the values of x, y and z. We have the following system.

$x+y =0 , 2x-3y=1, -2x+z=0$

We can solve for x like this x=-y and replacing into quation 2 we got:

$2(-y)-3y = 1$

$-5y = 1$

$y= -\frac{1}{5}$

And then we can solve for x and we got:

$x = -y = -(-\frac{1}{5})=\frac{1}{5}$

And if we solve for z we got:

$z=2x =2 \frac{1}{5}=\frac{2}{5}$

And now we can express the radius in terms of the dimensional analysis like this:

$r=K A^{1/5} \rho^{-1/5} t^{2/5}$

And K represent a constant in order to make the porportional relation and equality.

The problem says that we can assume the constant K=1.

And if we solve for the energy we got:

$A^{1/5}=\frac{r}{t^{2/5} \rho^{-1/5}}$

$A= \frac{r^5 \rho}{t^2}$

And now we can replace the values given. On this case t =0.025 s, the radius r =140 m, and the density is a constant assumed $\rho =1.2 kg/m^2$ , and replacing we got:

$A=\frac{140^5 1.2 kg/m^3}{(0.025 s)^2}=1.033x10^{14} \frac{kg m^2}{s^2}$

And we can convert this into ergs we got:

$A= 1.033x10^{14} \frac{kgm^2}{s^2} * \frac{1 x10^7 egrs}{1 \frac{kgm^2}{s^2}}=1.033x10^{21} ergs$

And then we know that 1 g of TNT have $4x10^4 erg$

And we got:

$A=1.033x10^{21} ergs *\frac{Kg TNT}{4x10^{10} erg}=2.58x10^{10} Kg TNT$

3 0

3 years ago

Why become an Android programmer?

Tpy6a [65]

As a free and open operating system, Android allows app developers to generate new ideas and work with a wide range of smartphones to open up hardware options, as companies supply a great variety of devices to choose from and high-end devices become more affordable.25 Jan 2017

4 0

3 years ago

You are performing a machining operation that approximates orthogonal cutting. Given that the chip thickness prior to chip forma

Mnenie [13.5K]

Answer:

shear plane angle = 31.98°

shear strain = cot ( 31.98° ) + tan ( 31.98 - 10 )

B) shear strength = 7339.78

Explanation:

a) Determine the shear plane angle and shear strain

Given data :

Chip thickness before chip formation = 0.5 inches

Chip thickness after separation = 1.125 inches

rake angle ( ∝ ) = 10°

shear plane angle : Tan ∅ = $\frac{rcos\alpha }{1-sin\alpha }$ ----- ( 1 )

r = chip thickness ratio = 0.5 / 1.125 = 0.4444

back to equation 1 : Tan ∅ = ( 0.444 ) * cos 10 / 1 - sin 10

Tan ∅ = 0.4444 * 0.9848 / 1 - 0.1736 = 0.5296

hence ∅ = tan^-1 ( 0.5296 ) = 31.98°

shear strain : R = cot ∅ + tan ( ∅ - ∝ ) ---------- ( 1 )

R = cot ( 31.98° ) + tan ( 31.98 - 10 )

B) determine the shear strength of the material

cutting force = 1559 N

thrust force = 1271 N

width of cut ( diameter ) = 3.0 mm

shear strength = c + σ.tan ∅

c = cohesion force = 1271 * 3 = 3813

σ = normal stress = F / A = 1559 / π/4 * ( 0.5 )^2 = 1559 / 0.1963 = 7941.94

hence : shear strength of material = 3813 + 7941.94 * 0.6244 = 7339.78

3 0

3 years ago