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Vaselesa [24]
3 years ago
6

Describe the are of mechanical engineering

Engineering
2 answers:
Alja [10]3 years ago
8 0
Bcc. Chilling. Vvcfgjkl
posledela3 years ago
5 0

Answer:

Mechanical engineering is an engineering discipline that combines engineering physics and mathematics principles with materials science to design, analyze, manufacture, and maintain mechanical systems.

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A well is located in a 20.1-m thick confined aquifer with a conductivity of 14.9 m/day and a storativity of 0.0051. If the well
ahrayia [7]

Answer:

S = 5.7209 M

Explanation:

Given data:

B = 20.1 m

conductivity ( K ) = 14.9 m/day

Storativity  ( s ) = 0.0051

1 gpm = 5.451 m^3/day

calculate the Transmissibility ( T ) = K * B

                                                       = 14.9 * 20.1 = 299.5  m^2/day

Note :

t = 1

U = ( r^2* S ) / (4*T*<em> t </em>)

  = ( 7^2 * 0.0051 ) / ( 4 * 299.5 * 1 ) = 2.0859 * 10^-4

Applying the thesis method

W(u) = -0.5772 - In(U)

       = 7.9

next we calculate the pumping rate from well ( Q ) in m^3/day

= 500 * 5.451 m^3 /day

= 2725.5 m^3 /day

Finally calculate the drawdown at a distance of 7.0 m form the well after 1 day of pumping

S = \frac{Q}{4\pi T} * W (u)

 where : Q = 2725.5

               T = 299.5

               W(u)  = 7.9

substitute the given values into equation above

S = 5.7209 M

4 0
3 years ago
Select all that apply: Contaminated sharps should<br><br> not be<br><br> ----
PIT_PIT [208]

Answer:

Contaminated sharps should not be bent, recapped or removed.

Explanation:

Contaminated sharps are defined as "any contaminated object that can penetrate the skin including, but not limited to, needles, scalpels, broken glass, broken capillary tubes and exposed ends of dental wires".

4 0
3 years ago
Air enters a diffuser operating at steady state at 540°R, 15 lbf/in.2, with a velocity of 600 ft/s, and exits with a velocity of
yKpoI14uk [10]

Answer: Hello the question is incomplete below is the missing part

Question:  determine the temperature, in °R, at the exit

answer:

T2= 569.62°R

Explanation:

T1 = 540°R

V2 = 600 ft/s

V1 = 60 ft/s

h1 = 129.0613  ( value gotten from Ideal gas property-air table )

<em>first step : calculate the value of h2 using the equation below </em>

assuming no work is done ( potential energy is ignored )

h2 = [ h1 + ( V2^2 - V1^2 ) / 2 ] * 1 / 32.2 * 1 / 778

∴ h2 = 136.17 Btu/Ibm

From Table A-17

we will apply interpolation

attached below is the remaining part of the solution

8 0
2 years ago
A beam has a rectangular cross section that is 5 inches wide and 1.5 inches tall. The supports are 60 inches apart and with a 12
nydimaria [60]

Answer:

The value of Modulus of elasticity E = 85.33 × 10^{6} \frac{lbm}{in^{2} }

Beam deflection is = 0.15 in

Explanation:

Given data

width = 5 in

Length = 60 in

Mass of the person = 125 lb

Load = 125 × 32 = 4000\frac{ft lbm}{s^{2} }

We know that moment of inertia is given as

I = \frac{bt^{3} }{12}

I = \frac{5 (1.5^{3} )}{12}

I = 1.40625 in^{4}

Deflection = 0.15 in

We know that deflection of the beam in this case is given as

Δ = \frac{PL^{3} }{48EI}

0.15 = \frac{4000(60)^{3} }{48 E (1.40625)}

E = 85.33 × 10^{6} \frac{lbm}{in^{2} }

This is the value of Modulus of elasticity.

Beam deflection is = 0.15 in

6 0
3 years ago
The cantilevered W530 x 150 beam shown is subjected to a 9.8-kN force F applied by means of a welded plate at A. Determine the e
snow_lady [41]

The <em>equivalent force-couple system</em> at O is the force and couple experienced when at point O due to the applied force at point A

The <em>equivalent force couple</em> system at O due to force <em>F</em> are;

Force, F =  (<u>8.65·i - 4.6·j</u>) KN

Couple, M₀ ≈ <u>40.9 </u>k kN·m

The reason the above values are correct is as follows:

The known values for the <em>cantilever</em> are;

The <em>height </em>of the beam = 0.65 m

The <em>magnitude of </em>the applied <em>force</em>, F = 9.8 kN

The <em>length </em>of the beam = 4.9 m

The <em>angle </em>away from the vertical the force is applied = 26°

The required parameter:

The <em>equivalent force-couple system</em> at the centroid of the beam cross-section of the cantilever

Solution:

The <em>equivalent force-couple system</em> is the force-couple system that can replace the given force at centroid of the beam cross-section at the cantilever O ;

The <em>equivalent force</em> \overset \longrightarrow F = 9.8 kN × cos(28°)·i - 9.8 kN × sin(28°)·j

Which gives;

The <em>equivalent force</em> \overset \longrightarrow F ≈ (<u>8.65·i - 4.6·j</u>) KN

The <em>couple </em><em>acting </em>at point O due to the force <em>F</em> is given as follows;

The <em>clockwise moment</em> = <em>9.8 kN × cos(28°) × 4.9</em>

The <em>anticlockwise moment</em> = <em>9.8 kN × sin(28°) 0.65/2 </em>

The sum of the moments = Anticlockwise moment - Clockwise moments

∴ The <em>sum </em>of the moments, ∑M, gives the moment acting at point O as follows;

M₀ = <em>9.8 kN × sin(28°) 0.65/2 - 9.8 kN × cos(28°) × 4.9</em>  ≈ 40.9 kN·m

The couple acting at O, due to F,  M₀ ≈ <u>40.9 kN·m</u>

The equivalent force couple system acting at point O due the force, F, is as follows

F =  (8.65·i - 4.6·j) KN

M₀ ≈ <u>40.9 </u>k kN·m

Learn more about equivalent force systems here:

brainly.com/question/12209585

4 0
3 years ago
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