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3 years ago

Please help me in this assignment.

1 answer:

8 0

<h2><em>Dell realizes that their ultimate success lies with the success of their supply chainand its ability to generate supply chain surplus. If Dell was to view supply chainoperations as a zero sum game, they would lose their competitive edge as their suppliers’ businesses struggled. Dell’s profit gained at the expense of their supplychain partners would be short lived. Just as a physical chain is only as strong as itsweakest link, the supply chain can be successful only if all members cooperateand focus on a global optimum rather than many local optima.</em></h2><h2><em></em></h2><h2><em></em></h2><h2><em>HOPE IT HELPS (◕‿◕✿)</em></h2>

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A pressure gage connected to a tank reads 50 psi at a location where the barometric reading is 29.1 inches Hg. Determine the abs

Effectus [21]

Answer:

Absolute pressure , P(abs)= 433.31 KPa

Explanation:

Given that

Gauge pressure P(gauge)= 50 psi

We know that barometer reads atmospheric pressure

Atmospheric pressure P(atm) = 29.1 inches of Hg

We know that

1 psi = 6.89 KPa

So 50 psi = 6.89 x 50 KPa

P(gauge)= 50 psi =344.72 KPa

We know that

1 inch = 0.0254 m

29.1 inches = 0.739 m

Atmospheric pressure P(atm) = 0.739 m of Hg

We know that density of Hg = $13.6\times 10^3\ kg/m^3$

P = ρ g h

P(atm) = 13.6 x 1000 x 9.81 x 0.739 Pa

P(atm) = 13.6 x 9.81 x 0.739 KPa

P(atm) =98.54 KPa

Now

Absolute pressure = Gauge pressure + Atmospheric pressure

P(abs)=P(gauge) + P(atm)

P(abs)= 344.72 KPa + 98.54 KPa

P(abs)= 433.31 KPa

3 0

3 years ago

What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 2.5×10-4

krok68 [10]

Answer:

1788.9 MPa

Explanation:

The magnitude of the maximum stress (σ) can be calculated usign the following equation:

$\sigma = 2\sigma_{0} \sqrt{\frac{a}{\rho}}$

<u>Where:</u>

<em>ρ: is the radius of curvature = 2.5x10⁻⁴ mm (0.9843x10⁻⁵ in)</em>

<em>σ₀: is the tensile stress = 100x10⁶ Pa (14500 psi) </em>

<em>2a: is the crack length = 4x10⁻² mm (1.575x10⁻³ in) </em>

Hence, the maximum stress (σ) is:

$\sigma = 2*100\cdot 10^{6} Pa \sqrt{\frac{(4 \cdot 10^{-2} mm)/2}{2.5 \cdot 10^{-4} mm}} = 1.79 \cdot 10^{6} Pa = 1788.9 MPa$

Therefore, the magnitude of the maximum stress is 1788.9 MPa.

I hope it helps you!

5 0

4 years ago

When trying to solve a frame problem it will typically be necessary to draw many free body diagrams. a)-True b)-False

klasskru [66]

Answer:

True

Explanation:

When trying to solve a frame problem in Engineering or Physics, it will typically be necessary to draw more than one body diagram.

When we have several parts of the frame or a set of frames, we have the anchor point, as well as the intersections of frames. Besides that, usually, there is a particle or rigid body together with the frame system. In this sense, usually, it is required to analyze a body diagram for the particle or rigid body suspended, as well as the intersections of the frames. So, usually, it will be required a minimum of two body diagrams.

If the system is more complex, or there are many intersections points, it will be required more than two body diagrams.

Finally, indeed, it will typically be necessary to draw many-body diagrams.

6 0

4 years ago

Using any of the bilinear transform, matched pole-zero, or impulse invariance techniques in converting a continuous-time system

leonid [27]

Answer:

A. True

The bilinear transform is employed in digital signal processing and discrete-time control theory which helps in transforming continuous-time system representations to discrete-time

4 0

3 years ago

Read 2 more answers

Very thin films are usually deposited under vacuum conditions to prevent contamination and ensure that atoms can fly directly fr

katrin [286]

Answer:

a. 9947 m

b. 99476 times

c. 2*10^11 molecules

Explanation:

a) To find the mean free path of the air molecules you use the following formula:

$\lambda=\frac{RT}{\sqrt{2}\pi d^2N_AP}$

R: ideal gas constant = 8.3144 Pam^3/mol K

P: pressure = 1.5*10^{-6} Pa

T: temperature = 300K

N_A: Avogadros' constant = 2.022*10^{23}molecules/mol

d: diameter of the particle = 0.25nm=0.25*10^-9m

By replacing all these values you obtain:

$\lambda=\frac{(8.3144 Pa m^3/mol K)(300K)}{\sqrt{2}\pi (0.25*10^{-9}m)^2(6.02*10^{23})(1.5*10^{-6}Pa)}=9947.62m$

b) If we assume that the molecule, at the average, is at the center of the chamber, the times the molecule will collide is:

$n_{collision}=\frac{9947.62m}{0.05m}\approx198952\ times$

c) By using the equation of the ideal gases you obtain:

$PV=NRT\\\\N=\frac{PV}{RT}=\frac{(1.5*10^{-6}Pa)(\frac{4}{3}\pi(0.05m)^3)}{(8.3144Pa\ m^3/mol\ K)(300K)}=3.14*10^{-13}mol\\\\n=(3.14*10^{-13})(6.02*10^{23})\ molecules\approx2*10^{11}\ molecules$

5 0

3 years ago