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____ [38]
3 years ago
5

How many color are in da rainbow​

Engineering
2 answers:
Juli2301 [7.4K]3 years ago
8 0

Answer:

The classic rainbow has 6 colours. However, I have heard it other ways. The classic is listed below, as well as one example of the various colour patterns used in rainbows.

Examples:

Classic: Red, Orange, Yellow, Green, Blue, Purple

Ex. 1: Red, Orange, Yellow, Green, Blue, Indigo, Violet

Good luck!!

abruzzese [7]3 years ago
5 0
Seven colors are in the rainbow haha
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The acrylic plastic rod is 200 mm long and 15 mm in diameter. If an axial load of 300 N is applied to it, determine the change i
ehidna [41]

Answer:

Change in length = 0.1257 mm

Change in diameter= -0.03771mm

Explanation:

Given

Diameter, d = 15 mm

Length of rod, L = 200mm

F = Force= 300N

d = 0.015m

Ep=2.70 GPa, np=0.4.

First, we have to calculate the normal stress using

σ = F/A where F = Force acting on the Cross-sectional area

A = Area

Area is calculated as πd²/4 where d = 0.015m

A = 22/7 * 0.015²/4

A = 0.000176785714285m²

A = 1.768E-4m²

So, stress. σ = 300N/1.768E-4m²

σ = 1696832.579185520Pa

σ = 1.697MPa

Calculating E(long)

E(long) = σ /Ep

E(long) = 1.697E-3/2.70

E(long) = 0.0006285

At this point, we fan now calculate the change in length of the element;

∆L = E(long) * L

∆L = 0.0006285 * 200mm

∆L = 0.1257mm

Calculating E(lat)

E(lat) = -np * E(long)

E(lat) = -4 * 0.0006285

E(lat) = -0.002514

At this point, we can now calculate the change in diameter of the element;

∆D = E(lat) * D

∆L = -0.002514 * 15mm

∆L = -0.03771mm

8 0
3 years ago
Carbon monoxide coming from your car smells like
Lemur [1.5K]
I wouldn’t smell like anything carbon monoxide is an odorless gas
3 0
3 years ago
The magic square is an arrangement of numbers in a square grid in such a way that the sum of the numbers in each row, and in eac
Yuki888 [10]

Answer:

See the explanation for the answer;

Explanation:

Matlab code is as given below

-------------------------------------------------------------------------------------Start of code

% Program to create a magic square and verify it

clc

M= magic(5)

% To find sum of elements of each row

r1= sum(M(1,:))      % Sum of row 1

r2= sum(M(2,:))      % Sum of row 2

r3= sum(M(3,:))      % Sum of row 3

r4= sum(M(4,:))      % Sum of row 4

r5= sum(M(5,:))      % Sum of row 5

% To find sum of each coloumn

c1= sum(M(:,1))      % Sum of coloumn 1

c2= sum(M(:,2))      % Sum of coloumn 2

c2= sum(M(:,3))      % Sum of coloumn 3

c2= sum(M(:,4))      % Sum of coloumn 4

c2= sum(M(:,5))      % Sum of coloumn 5

% To find sum of diagonal

d1= sum(diag(M))               % Sum of principal diagonal elements

d2= sum(diag(fliplr(M)))      

-------------------------------------------------------------------------------End of code

Following results are obtained when executed.

M =

   17    24     1     8    15

   23     5     7    14    16

    4     6    13    20    22

   10    12    19    21     3

   11    18    25     2     9

r1 =

   65

r2 =

   65

r3 =

   65

r4 =

   65

r5 =

   65

c1 =

   65

c2 =

   65

c2 =

   65

c2 =

   65

c2 =

   65

d1 =

   65

d2 =

   65

3 0
3 years ago
A zener diode exhibits a constant voltage of 5.6 V for currents greater than five times the knee current. IZK is specified to be
8090 [49]

Answer:

The maximum power dissipation of the zener diode 112mV.

Explanation:

The minimum zener current should be:

5 * Iza= 5 * 1=  5 mA.

Since the load current can be at maximum 15 mA, we should select R so that, IL= 15 mA.

A zener current of 5 mA is available, Thus the current should be 20 mA, which leads to,

R = \frac{15 - 5.6}{20 mA} = 470 Ω.

Maximum power dissipated in the diode occours when, IL=0 is

Pmax = 20 * 10^{3} * 5.6 = 112mV.

5 0
4 years ago
Carbon dating for archeological materials is based on the fact that a plant, after its death, stops absorbing radioactive C-14 a
Olin [163]

Answer:

 t = 2212 years

Explanation:

In radioactive decay processes it is described by the equation

         N = N₀ e^{-\lambda t}

to calculate the activity

        T_{1/2} = log 2 /λ

        λ = log 2 / T_{1/2}

     

        λ = log 2 /5715

        λ = 5.267 10⁻⁵

now the amount of carbon 14 is N₀ = 0.1%, the sample contains an amount of N = 0.089%

          N / N₀ = e^{-\lambda t}

          -λ t = ln N / N₀

           t = - 1 /λ  ln N /N₀

           t = 1 / 5.267 10⁻⁵   ln (0.089 / 0.1)

           t = 2,212 10³ years

           t = 2212 years

8 0
3 years ago
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