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3 years ago

Documentation of a flow chart?

Engineering

1 answer:

jonny [76]3 years ago

8 0

Answer:

A Document in a flowchart represents a printout, for instance a document or a report.

Flowcharts are used in planning and documenting basic process or projects. Like different sorts of outlines, they help imagine what is happening and along these lines help comprehend a process, and maybe additionally find more subtle highlights inside the process.

Explanation:

A flowchart is essentially a graphical portrayal of steps. It shows steps in consecutive request and is generally used in introducing the progression of calculations or work process. Regularly, a flowchart shows the means as boxes of different sorts, and their request by interfacing them with bolts.

It is likewise a kind of graph that speaks to a work process or cycle. A flowchart can likewise be characterized as a diagrammatic portrayal of a calculation, a bit by bit way to deal with settling an assignment.

The flowchart shows the means as boxes of different sorts, and their request by associating the containers with bolts. This diagrammatic portrayal outlines an answer model to a given issue. Flowcharts are used in investigating, planning, reporting or dealing with a process or program in different fields.

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Hello, I have a question, I would be glad if you can help.

Anastasy [175]

Answer:

infinite

Explanation:

The tangent of the angle the arm makes with the vertical will be the ratio of the centripetal acceleration to the acceleration due to gravity on the center of mass of the arm. The angle can only be 90° (fully-open arms) if that ratio is infinite.

The speed must be infinite for the arms to be fully open.

3 0

2 years ago

A steam turbine receives steam at 1.5MPa and 220oC, and exhausts at 50kPa, 0.75 dry. Neglecting heat losses and changes in kinet

SIZIF [17.4K]

Answer:

Can you make friend with me ?

4 0

3 years ago

(35-39) A student travels on a school bus in the middle of winter from home to school. The school bus temperature is 68.0° F. Th

arlik [135]

Answer:

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

Explanation:

From Heat Transfer we determine that heat transfer rate due to electromagnetic radiation ( $\dot Q$ ), measured in BTU per hour, is represented by this formula:

$\dot Q = \epsilon\cdot A\cdot \sigma \cdot (T_{s}^{4}-T_{b}^{4})$ (1)

Where:

$\epsilon$ - Emissivity, dimensionless.

$A$ - Surface area of the student, measured in square feet.

$\sigma$ - Stefan-Boltzmann constant, measured in BTU per hour-square feet-quartic Rankine.

$T_{s}$ - Temperature of the student, measured in Rankine.

$T_{b}$ - Temperature of the bus, measured in Rankine.

If we know that $\epsilon = 0.90$ , $A = 16.188\,ft^{2}$ , $\sigma = 1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}}$ , $T_{s} = 554.07\,R$ and $T_{b} = 527.67\,R$ , then the heat transfer rate due to electromagnetic radiation is:

$\dot Q = (0.90)\cdot (16.188\,ft^{2})\cdot \left(1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}} \right)\cdot [(554.07\,R)^{4}-(527.67\,R)^{4}]$

$\dot Q = 417.492\,\frac{BTU}{h}$

Under the consideration of steady heat transfer we find that the net energy transfer from the student's body during the 20 min-ride to school is:

$Q = \dot Q \cdot \Delta t$ (2)

Where $\Delta t$ is the heat transfer time, measured in hours.

If we know that $\dot Q = 417.492\,\frac{BTU}{h}$ and $\Delta t = \frac{1}{3}\,h$ , then the net energy transfer is:

$Q = \left(417.492\,\frac{BTU}{h} \right)\cdot \left(\frac{1}{3}\,h \right)$

$Q = 139.164\,BTU$

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

7 0

3 years ago

Two steel plates are to be held together by means of 16-mm-diameter high-strength steel bolts fitting snugly inside cylindrical

dusya [7]

Answer:

The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm

Explanation:

For steel bolt

Stress = 210 MPa or 210 N/mm2

Pressure = Stress* Area

Pbolt = 210 N/mm2 * 16^2 *(pi)/4

Pbolt = 210 N/mm2 * 200.96 mm^2 = 42201.6 N

For Brass spacer

Pressure = 42201.6 N

Area of Brass spacer = Pressure/Stress

Area of Brass spacer = 42201.6 N/145 N/mm^2 = 291.044 mm^2

Area of Brass spacer = (pi) (d^2 - 16^2)/4 = 291.044 mm^2

d^2 - 16^2 = 291.044 mm^2* 4/(pi) = 370.758

d^2 = 370.758 + 16^2

d^2 = 626.758

d = 25.03 mm

The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm

5 0

3 years ago

Can someone explain it to me!

Darina [25.2K]

Answer:

It will be B

Explanation:

Since resisitors in series are added together, 1 + 1 + 1 would = 3kilo ohms. But with resistors in parallel would be (1/1+1/1)^-1. That would equal 0.5 Now you have two resistors in series for B, and because now that they are in series you add them together, so 0.5 + 1 = 1.5 kilo ohms which is what is needed.

7 0

2 years ago