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valentinak56 [21]
3 years ago
11

When preparing a foundation for a heavy duty machine tool, discuss any four (4) statics machine characteristics to be considered

.
Engineering
1 answer:
kozerog [31]3 years ago
5 0

Answer:

1 ) Accuracy of the Machine Tool

2) Load bearing capacity

3) Linearity in the product line

4) Torque of the machine

Explanation:

we know that machine tool is the permanent essential in manufacturing industries

it is a machine use for different form like cutting , grinding and boring etc

so 1st is

1 ) Accuracy of the Machine Tool

we know it is very important Characteristic of the machine tool because when we use it in manufacturing unit Accuracy of the Machine Tool should be higher concern

2) Load bearing capacity

we should very careful about Load bearing capacity because how much amount of load tool will bear check by some parameter like creep , shear stress and strength etc

3) Linearity in the product line

Linearity in the product line mean that it should be group of related product produced by the any one of the manufacturer otherwise it will take time or it may be intermixing

4) Torque of the machine

we know that Torque is a rotational force or a turning force so amount of force multiplied by the distance of the operation

and we know torque per second give the power rating of machine tool

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vertical gate in an irrigation canal holds back 12.2 m of water. Find the average force on the gate if its width is 3.60 m. Repo
DanielleElmas [232]

Answer:

The right solution is "2625 kN".

Explanation:

According to the question,

The average pressure will be:

= density\times g\times \frac{h}{2}

By putting values, we get

= 1000\times 9.8\times \frac{12.2}{2}

= 1000\times 9.8\times 6.1

= 59780

hence,

The average force will be:

= Pressure\times Area

= 59780\times 3.6\times 12.2

= 2625537 \ N

Or,

= 2625 \ kN

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2 years ago
Can someone draw a electric circuit with electrical symbols for this
erastovalidia [21]

Answer:

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Explanation:

4 0
3 years ago
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Two uniformly charged conducting plates are parallel to each other. They each have area A. Plate #1 has a positive charge Q whil
Karo-lina-s [1.5K]

Answer:

E = \frac{3Q}{2A\epsilon_0}

Explanation:

By Gauss Law for electric field:

E = \frac{\sigma}{2\epsilon_0}

Where \sigma is the charge density Q/A. Since we have 2 parallel  plates with different charge, the electric field at point P in the gap would be the sum of 2 field

E = E_1 + E_2

E = \frac{Q}{2A\epsilon_0} + \frac{2Q}{2A\epsilon_0}

E = \frac{3Q}{2A\epsilon_0}

5 0
3 years ago
Whats the boolean expression of this circuit?
Natalija [7]

Answer:

  G8 = x0'x2' +x0'x3' +x1x2

Explanation:

The expression can be written different ways, depending on the need to avoid hazards. One of them is ...

  G_8=\overline{X_0}\,\overline{X_2}+\overline{X_0}\,\overline{X_3}+X_1X_2

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A truth table and Karnaugh map are shown for the circuit. The terms used in the Boolean expression come from the corners, the upper half of the left- and right-columns, and the right half of the middle two rows. If a static hazard is to be avoided, a term x1x0' could be added representing the right column.

8 0
2 years ago
A cantilever timber beam with a span of L = 4.25 m supports a linearly distributed load with maximum intensity of w0 = 5.5 kN/m.
9966 [12]

Answer:

the minimum width is b= 0.1414m = 141mm

Explanation:

]given,

L= 4.25

w₀ = 5.5kN/m,

allowable bending stress = 7MPa

allowable shear stress = 875kPa

h/b = 0.67

b = ?

for a linearly distributed load, with maximum load intensity, w₀ of 5.5kN/m,

the maximum moment, M exerted by the timber is =  \frac{w₀ L²}{9√3}[/texM = w₀ L²}/{9√3 = 99.34/15.6 =6.367kNmfor a linearly distributed load, with maximum load intensity, w₀ of 5.5kN/m, the shear force, V =  [tex]\frac{w₀ L}{2}[/texV = {w₀ L}/{3} = 7.79kNfor maximum bending stress of a rectangular timber, B, = [tex]\frac{6M}{bh²}

given h/b = 0.67, i.e h=0.67b

allowable bending stress = \frac{6M}{bh²} = 7000kPa

7000  = (6*6.37)/ (b *(0.67b)² ) = 38.21/0.449b³

3080b³=38.21

b³ = 38.21/3080 = 0.0124

b = 0.232m

h=0.67b = 0.67* 0.232 = 0.155m

for allowable  shear stress = (3V)/(2bh)

875 = (3V)/(2bh) = (3x 7.79)/(2xbx0.67b)

875 = 23.375/1.34b²

1172.5 b²= 23.375

b² =0.0199

b= 0.1414m

h=0.67b = 0.67* 0.1414 = 0.095m

the minimum width is b= 0.1414m = 141mm

4 0
3 years ago
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