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sergeinik [125]
3 years ago
9

The set of frequencies of the electromagnetic waves emitted by the atoms of an element is called

Physics
1 answer:
VLD [36.1K]3 years ago
7 0

Answer:

The set of frequencies of the electromagnetic Waves emitted by the atoms of an element is called emission spectrum.

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Which of the following is not considered a major work flow structure?
mixas84 [53]

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.

8 0
3 years ago
I need help with the circled one
Whitepunk [10]
Yes the answer is yes 
7 0
3 years ago
An insect 5.25 mm tall is placed 25.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the righ
Zigmanuir [339]

Answer:

(A) therefore the image is

  • 63 cm to the right of the lens
  • the image size is -13.22 cm
  • it is real
  • it is inverted

(B) therefore the image is

  • 63 cm to the right of the lens
  • the image size is -13.22 cm
  • it is real
  • it is inverted

Explanation:

height of the insect (h) = 5.25 mm = 0.525 cm

distance of the insect (s) = 25 cm

radius of curvature of the flat left surface (R1) = ∞

radius of curvature of the right surface (R2) = -12.5 cm (because it is a planoconvex lens with the radius in the direction of the incident rays)

index of refraction (n) = 1.7

(A) we can find the location of the image by applying the formula below

\frac{1}{f} =\frac{1}{s'} +\frac{1}{s} where

  • s' = distance of the image
  • f = focal length
  • but we first need to find the focal length before we can apply this formula

\frac{1}{f} =(n-1)(\frac{1}{R1} -\frac{1}{R2} )

\frac{1}{f} =(1.7-1)(\frac{1}{∞} -\frac{1}{-12.5} )

\frac{1}{f} =(0.7)(0 + \frac{1}{12.5} )

\frac{1}{f} =\frac{0.7}{12.5}

f = \frac{12.5}{0.7}

f = 17.9 cm

now that we have the focal length we can apply \frac{1}{f} =\frac{1}{s'} +\frac{1}{s}

\frac{1}{f} - \frac{1}{s} =\frac{1}{s'}

\frac{1}{17.9} - \frac{1}{25} =\frac{1}{s'}

\frac{25 - 17.9}{17.9 x 25} =\frac{1}{s'}

\frac{7.1}{447.5} =\frac{1}{s'}

s' = \frac{447.5}{7.1}[/tex]  = 63 cm to the right of the lens

magnification =\frac{-s'}{s} =\frac{y'}{y}   where y' is the height of the image, therefore

\frac{-s'}{s} =\frac{y'}{y}

\frac{-63}{25} =\frac{y'}{52.5}

y' = \frac{-63}{25} x 0.525 = -13.22 cm

therefore the image is

  • 63 cm to the right of the lens
  • the image size is -13.22 cm
  • it is real
  • it is inverted

(B) if the lens is reversed, the radius of curvatures would be interchanged

radius of curvature of the flat left surface (R1) = ∞

radius of curvature of the right surface (R2) = 12.5 cm

we can find the location of the image by applying the formula below

\frac{1}{f} =\frac{1}{s'} +\frac{1}{s} where

  • s' = distance of the image
  • f = focal length
  • but we first need to find the focal length before we can apply this formula

\frac{1}{f} =(n-1)(\frac{1}{R1} -\frac{1}{R2} )

\frac{1}{f} =(1.7-1)(\frac{1}{12.5} -\frac{1}{∞} )

\frac{1}{f} =(0.7)( \frac{1}{12.5} - 0)

\frac{1}{f} =\frac{0.7}{12.5}

f = \frac{12.5}{0.7}

f = 17.9 cm

now that we have the focal length we can apply \frac{1}{f} =\frac{1}{s'} +\frac{1}{s}

\frac{1}{f} - \frac{1}{s} =\frac{1}{s'}

\frac{1}{17.9} - \frac{1}{25} =\frac{1}{s'}

\frac{25 - 17.9}{17.9 x 25} =\frac{1}{s'}

\frac{7.1}{447.5} =\frac{1}{s'}

s' = \frac{447.5}{7.1}[/tex]  = 63 cm to the right of the lens

magnification =\frac{-s'}{s} =\frac{y'}{y}   where y' is the height of the image, therefore

\frac{-s'}{s} =\frac{y'}{y}

\frac{-63}{25} =\frac{y'}{52.5}

y' = \frac{-63}{25} x 0.525 = -13.22 cm

therefore the image is

  • 63 cm to the right of the lens
  • the image size is -13.22 cm
  • it is real
  • it is inverted

7 0
3 years ago
A mass accelerates uniformly when the resultant force on it:
WARRIOR [948]
F = ma
So if you want the "a" to stay constant (as it is then uniform acceleration), the F must also be constant to apply a constant force.

2 is the answer

3 0
3 years ago
A charged particle A exerts a force of 2.45 μN to the right on charged particle B when the particles are 12.2 mm apart. Particle
Brilliant_brown [7]

Answer:

F_2 = 1.10 \mu N

Explanation:

As we know that the electrostatic force is a based upon inverse square law

so we have

F = \frac{kq_1q_2}{r^2}

now since it depends inverse on the square of the distance so we can say

\frac{F_1}{F_2} = \frac{r_2^2}{r_1^2}

now we know that

r_2 = 18.2 mm

r_1 = 12.2 mm

also we know that

F_1 = 2.45 \mu N

now from above equation we have

F_2 = \frac{r_1^2}{r_2^2} F_1

F_2 = \frac{12.2^2}{18.2^2}(2.45\mu N)

F_2 = 1.10 \mu N

5 0
3 years ago
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