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2 years ago

Which of the following is not considered a major work flow structure?

Physics

1 answer:

mixas84 [53]2 years ago

8 0

Answer:

The correct option is;

D. Fabrication

Explanation:

A workflow flow is a detailed business process consisting of a series of required interconnected tasks in directed graph format that is executable by workflow management system.

Considering each of the options, we have

A. Work center

This consists of part of the transformation input to output. The location

B. Project

This is the unique identifier of the task to be processed

C. Assembly line

Forms part of the required input where transformation takes place and items are being processed within the assembly line

D. Fabrication

Here the item is fixed, without motion, therefore this is not considered a major work flow structure

E. Continuous flow

Here again, the items are being processed and are in motion, which constitutes a workflow structure.

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An isotope of the element fluorine has 9 protons and 10 neutrons. What is the name of this isotope? fluorine-

Tasya [4]

The name is fluorine-19.

Usually, in the names of isotopes, we say the name of the element followed the mass number (eg. Carbon-12). The mass number is the sum of the number of protons and neutrons of that atom. Electrons are not included since they're too light compared to protons and neutrons.

Therefore, to find out the name of the isotope, we have to find the mass number. Add up 9 and 10, which makes 19, so, the answer is fluorine-19.

5 0

3 years ago

Read 2 more answers

A force in the +x -direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 7.90 kg box that is sitting on the horizontal

dsp73

Answer:

$v\approx 8.570\,\frac{m}{s}$

Explanation:

The equation of equlibrium for the box is:

$\Sigma F_{x} = 18\,N-(0.530\,\frac{N}{m} )\cdot x = (7.90\,kg)\cdot a$

The formula for the acceleration, given in $\frac{m}{s^{2}}$ , is:

$a = \frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg}$

Velocity can be derived from the following definition of acceleration:

$a = v\cdot \frac{dv}{dx}$

$v\, dv = a\, dx$

$\frac{1}{2}\cdot v^{2} = \int\limits^{17\,m}_{0\,m} {\frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg} } \, dx$

$\frac{1}{2}\cdot v^{2} =\frac{18\,N}{7.90\,kg} \int\limits^{17\,m}_{0\,m}\, dx - \frac{0.530\,\frac{N}{m} }{7.90\,kg} \int\limits^{17\,m}_{0\,m} {x} \, dx$

$\frac{1}{2}\cdot v^{2} = (2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}$

$v =\sqrt{2\cdot[(2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}] }$

The speed after the box has travelled 17 meters is:

$v\approx 8.570\,\frac{m}{s}$

3 0

3 years ago

Raindrops fall vertically at 7.5 m/s relative to the Earth. What does an observer in a car moving at 20.2 m/s in a straight line

Vilka [71]

Answer:

vDP = 21.7454 m/s

θ = 200.3693°

Explanation:

Given

vDE = 7.5 m/s

vPE = 20.2 m/s

Required: vDP

Assume that

vDE to be in direction of - j

vPE to be in direction of i

According to relative motion concept the velocity vDP is given by

vDP = vDE - vPE (I)

Substitute in (I) to get that

vDP = - 7.5 j - 20.2 i

The magnitude of vDP is given by

vDP = √((- 7.5)²+(- 20.2)²) m/s = 21.7454 m/s

θ = Arctan (- 7.5/- 20.2) = 20.3693°

θ is in 3rd quadrant so add 180°

θ = 20.3693° + 180° = 200.3693°

4 0

3 years ago

G to ml,density=3.291 g/ml

olya-2409 [2.1K]

The density value 3.291 g/ml (gram / milliliter) in words is "three point two nine one g/ml (gram / milliliter)".
=)

7 0

2 years ago

A force of 720 Newton stretches a spring 4 meters. A mass of 45 Kilograms is attached to the spring and is initially released fr

poizon [28]

Answer:

x(t) = -3sin2t

Explanation:

Given that

Spring force of, W = 720 N

Extension of the spring, s = 4 m

Attached mass to the spring, m = 45 kg

Velocity of, v = 6 m/s

The proper calculation is attached via the image below.

Final solution is x(t) = -3.sin2t

5 0

3 years ago