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Illusion [34]
3 years ago
5

Identify SIX (6) objectives of maintenance.​

Engineering
1 answer:
Rasek [7]3 years ago
8 0

Answer:

to optimize the reliability of equipment and infrastructure;

- to ensure that equipment and infrastructure are always in good condition;

- to carry out prompt emergency repair of equipment and infrastructure so as to secure the best possible availability for production;

- to enhance, through modifications, extensions, or new low-cost items, the productivity of existing equipment or production capacity;

- to ensure the operation of equipment for production and for the distribution of energy and fluids;

- to improve operational safety;

- to train personnel in specific maintenance skills;

- to advise on the acquisition, installation and operation of machinery;

- to contribute to finished product quality;

- to ensure environmental protection.

Explanation:

pick whichever you want

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A projectile is launched horizontally 1m above the ground. If it lands 300m away from the initial launch position, find: a)-the
PIT_PIT [208]

Answer:

(a): The launch velocity is Vx= 666.66 m/s.

(b): The angle wich the projectile contacts the ground is α= 0.38°

Explanation:

h= 1m

g= 9.8 m/s²

h= g*t²/2

t= 0.45 s

Vy= g*t

Vy= 4.42 m/s

d=Vx* t

Vx= 666.66 m/s (a)

α= tg⁻¹ ( Vy/Vx)

α= 0.38° (b)

4 0
3 years ago
I am trying to create a line of code to calculate distance between two points. (distance=[tex]\sqrt{ (x2-x1)^2+(y2-y1)^2}) My li
k0ka [10]

Answer:

point_dist = math.sqrt((math.pow(x2 - x1, 2) + math.pow(y2 - y1, 2))

Explanation:

The distance formula is the difference of the x coordinates squared, plus the difference of the y coordinates squared, all square rooted.  For the general case, it appears you simply need to change how you have written the code.

point_dist = math.sqrt((math.pow(x2 - x1, 2) + math.pow(y2 - y1, 2))

Note, by moving the 2 inside of the pow function, you have provided the second argument that it is requesting.

You were close with your initial attempt, you just had a parenthesis after x1 and y1 when you should not have.

Cheers.

6 0
3 years ago
What is one of the “don’ts” in drawing dimension lines? they should never be labeled they should never be stacked they should ne
Strike441 [17]

Answer:

What is one of the “don’ts” in drawing dimension lines? they should never be labeled they should never be stacked they should never cross each other they should never have only one measurement value

5 0
4 years ago
Read 2 more answers
A diesel engine with CR= 20 has inlet at 520R, a maximum pressure of 920 psia and maximum temperature of 3200 R. With cold air p
Stella [2.4K]

Answer:

Cut-off ratio\dfrac{V_3}{V_2}=6.15

Cxpansion ratio\dfrac{V_4}{V_3}=3.25

The exhaust temperatureT_4=1997.5R

Explanation:

Compression ratio CR(r)=20

\dfrac{V_1}{V_2}=20

P_2=P_3=920 psia

T_1=520 R ,T_{max}=T_3,T_3=3200 R

We know that for air γ=1.4

If we assume that in diesel engine all process is adiabatic then

\dfrac{T_2}{T_1}=r^{\gamma -1}

\dfrac{T_2}{520}=20^{1.4 -1}

T_2=1723.28R

\dfrac{V_3}{V_2}=\dfrac{T_3}{T_2}

\dfrac{V_3}{V_2}=\dfrac{3200}{520}

So cut-off ratio\dfrac{V_3}{V_2}=6.15

\dfrac{V_1}{V_2}=\dfrac{V_4}{V_3}\times\dfrac{V_3}{V_2}

Now putting the values in above equation

\dfrac20=\dfrac{V_4}{V_3}\times 6.15

\dfrac{V_4}{V_3}=3.25

So expansion ratio\dfrac{V_4}{V_3}=3.25.

\dfrac{T_4}{T_3}=(expansion\ ratio)^{\gamma -1}

\dfrac{T_3}{T_4}=(3.25)^{1.4 -1}

T_4=1997.5R

So the exhaust temperatureT_4=1997.5R

3 0
4 years ago
A 1-ft rod with a diameter of 0.5 in. is subjected to a tensile force of 1,300 lb and has an elongation of 0.009 in. The modulus
iragen [17]

Answer:

E = 8.83 kips

Explanation:

First, we determine the stress on the rod:

\sigma = \frac{F}{A}\\\\

where,

σ = stress = ?

F = Force Applied = 1300 lb

A = Cross-sectional Area of rod = 0.5\pi \frac{d^2}{4} = \pi \frac{(0.5\ in)^2}{4} = 0.1963\ in^2

Therefore,

\sigma = \frac{1300\ lb}{0.1963\ in^2} \\\\\sigma = 6.62\ kips

Now, we determine the strain:

strain = \epsilon = \frac{elongation}{original\ length} \\\\\epsilon = \frac{0.009\ in}{12\ in}\\\\\epsilon =  7.5\ x\ 10^{-4}

Now, the modulus of elasticity (E) is given as:

E = \frac{\sigma}{\epsilon}\\\\E = \frac{6.62\ kips}{7.5\ x\ 10^{-4}}

<u>E = 8.83 kips</u>

7 0
3 years ago
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