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valentinak56 [21]
4 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]4 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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Ugo [173]

Answer:

an equivalent weight or force : counterbalance.

4 0
3 years ago
Read 2 more answers
what is the relationship between the voltages of the output and adjust pins of LM1117 when in normal operation
Elza [17]

Based on the engineering analysis, the relationship between the voltages of the output and adjust pins of LM1117 when in normal operation is "the LM1117 adjustable version establishes a 1.25V reference voltage, (VREF) between the output and the adjust terminal."

The LM1117 generally has a 1.2V at 800mA of load current.

The LM1117 is known to be a progression of low dropout linear voltage regulators. It is adjustable and be assigned between the output voltage of 1.25 to 13.8 V using just two external resistors.

Hence, in this case, it is concluded that the correct answer is "the LM1117 adjustable version establishes a 1.25V reference voltage, (VREF), between the output and the adjust terminal.

Learn more here: brainly.com/question/25646477

4 0
3 years ago
If the open circuit voltage of a circuit containing ideal sources and resistors is measured at 6 V, while the current through th
Trava [24]

Given Information:

Current of Ideal source = Is = 100 mA = 0.100 A

Open circuit voltage = Voc = 6 V

Short circuit current = Isc = 200 mA = 0.200 A

Required Information:

Power absorbed by idea current source = ?

Answer:

Power absorbed by idea current source = 0.3 Watts

Explanation:

Ideal Current Source:

An ideal current source doesn't have internal resistance and provides constant current regardless of the voltage supplied to the circuit.

The power absorbed by the ideal source can be found using

P = Is²R

Where Is is the ideal source current and R can be found using

R = Voc/Isc

Where Voc is the open circuit voltage and Isc is the short circuit current.

R = 6/0.200

R = 30 Ω

Therefore, the power absorbed by the ideal current source is

P = Is²R

P = (0.100)²*30

P = 0.3 Watts

7 0
3 years ago
A tank has a gauge pressure of 552 psi. The cover of an inspection port on the tank has a surface area of 18 square inches. What
Reptile [31]

Answer:

44197.55 N

Explanation:

From the question,

Pressure of the pressure guage (P) = Total force experienced by the cover (F)/Area of the cover (A)

P = F/A................ Equation 1

make F the subeject of the equation

F = P×A............... Equation 2

Given: P = 552 psi = (552×6894.76) = 3805907.52 N/m², A = 18 square inches = (18×0.00064516) = 0.01161288 m²

Substitute these values into equation 2

F = ( 3805907.52×0.01161288)

F = 44197.55 N

3 0
3 years ago
Consider three charged sheets, A, B, and C. The sheets are parallel with A above B above C. On each sheet there is surface charg
babunello [35]

Answer:

0.

Explanation:

To find the electrical force per unit area on each sheet we start defining our variables,

\sigma_A = -4.10^{-5}C/m^2

\sigma_B= -7.10^{-5}C/m^2

\sigma_C = -3.1^{-5}C/m^2

We find the electric field for each one, this formula is given by,

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

Substituting each value from the three charged sheets, we have

E_A= \frac{-4.10^{-5}C/m^2}{2\epsilon_0}(\hat{j})

E_B= \frac{-7.10^{-5}C/m^2}{2\epsilon_0}(-\hat{j})

E_C= \frac{-3.1^{-5}C/m^2}{2\epsilon_0}(\hat{j})

The electric field is

E_{NET}= E_A+E_B+E_C

E_{NET}=\frac{-4.10^{-5}C/m^2}{2\epsilon_0}(\hat{j})+\frac{-7.10^{-5}C/m^2}{2\epsilon_0}(-\hat{j})+ \frac{-3.1^{-5}C/m^2}{2\epsilon_0}(\hat{j})

E_{NET} = 0

Force on each sheet is,

F=E_{NET}\sigma ds

F=0

The total force is 0

5 0
3 years ago
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