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galina1969 [7]
3 years ago
9

A scientist makes observations of a forest environment. Describe an example of a pattern that the scientist might’ve observe

Engineering
2 answers:
krek1111 [17]3 years ago
5 0

Answer:

The answer is different types of animals and plants, and the forest soil.

Explanation:

A forest Ecosystems consist of different animals and plants found there as well as the soil that in-habitat several microorganisms. While explaining the details of forest ecosystem is it important to describe that what kind of animals are found there and what is the role of forest soil in the growth of plantation found there.

ZanzabumX [31]3 years ago
5 0

Answer:

Scientists observe in many ways – with their own senses or with tools such as microscopes, scanners or transmitters to extend their vision or hearing. These tools allow for more precise and accurate observations. Scientists also use equipment to measure things like radiation or pH – phenomena not directly observable.

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13. Write a function which is passed two strings. The function creates a new string from the two original strings by copying one
attashe74 [19]

Answer:

I am writing the code in C++. Let me know if you want the program in some other programming language.

#include <iostream>  // includes header file for input output functions

using namespace std;     //to identify objects like cin cout

string CopyStrings(string string1, string string2)  

{   string newString = "";    

   for (int loop = 0; loop < string1.length() ||  

                   loop < string2.length(); loop++)      {      

       if (loop < string1.length())  

           newString += string1[loop];          

       if (loop < string2.length())  

           newString += string2[loop];      }  

   return newString;   }  

int main()  

{   string stringA = "ace";  

   string stringB = "bdf";  

   cout << CopyStrings(stringA, stringB);   }

Output:

abcdef

Explanation:

The function CopyStrings() function takes two strings i.e. string1 and string2 as parameters to copy characters from both the string one character from each.

The newString variable stores the new string after copying characters from both strings string1 and string2.

Then the for loop starts which has a variable loop which is an index variable that traverses through both the strings stored in string1 and string2. The loop continues to execute until it moves through entire length of string1 and string2 which means it copies all the characters from both string1 and string2. length() is used here which returns length of the string1 and string2.

If statement in the for loop checks the character that loop (index) variable is pointing to is less than the string1 length which means it checks each character stored in string1. For example if string1 contains "ace" and loop variable is moving through the string and is currently at "a" then this condition is true. If the condition evaluates to true then the body of if statement is executed. The next statement stores that character a into the newString variable.

Next If statement checks character that loop variable is pointing to is less than the string2 length which means it checks each character stored in string2. For example if string2 contains "bdf" and loop variable is moving through the string and is currently at "b" then this condition is true. If the condition evaluates to true then the body of if statement is executed. The next statement stores that character b into the newString variable.

Then the second iteration starts which again first stores the next character i.e. c from string1 into newString and then stores next character i.e d from string2 into newString.

Then the third iteration starts which again first stores the next character i.e. e from string1 into newString and then stores next character i.e f from string2 into newString.

Then the loop breaks as the loop variable reaches end of both the string1 and string2.

return newString will return the copied string into the output screen which is abcdef.

The screenshot of code along with output is attached.

3 0
3 years ago
Nêu đặc điểm của tín hiệu PAM rời rạc dạng lưỡng cực NRZ, RZ
zhuklara [117]

Answer:

yes it is certainly good ice cream

4 0
3 years ago
Let be a real-valued signal for which when . Amplitude modulation is preformed to produce the signal . A proposed demodulation t
densk [106]

Answer:

hello your question is incomplete attached below is the complete question

answer : attached below

Explanation:

let ; x(t)  be a real value signal for x ( jw ) = 0 , |w| > 200\pi

g(t) = x ( t ) sin ( 2000 \pi t )

x_{1} (t) = \frac{1}{2}  x(t)  sin ( 4000\pi t )

next we apply Fourier transform

attached below is the remaining part of the solution

6 0
3 years ago
What oil specification is used to express the thickness (viscosity) of the oil?​
Marianna [84]

Answer:

Viscosity is notated using the common classification “XW-XX”. The number preceding the “W” (winter) rates the oil's flow (viscosity) at zero degrees Fahrenheit (-17.8 degrees Celsius). The lower the number, the less the oil thickens in cold weather.

7 0
3 years ago
A pipeline (NPS = 14 in; schedule = 80) has a length of 200 m. Water (15℃) is flowing at 0.16 m3/s. What is the pipe head loss f
dangina [55]

Answer:

Head loss is 1.64

Explanation:

Given data:

Length (L) = 200 m

Discharge (Q) = 0.16 m3/s

According to table of nominal pipe size , for schedule 80 , NPS 14,  pipe has diameter (D)= 12.5 in or 31.8 cm 0.318 m

We know, head\ loss  = \frac{f L V^2}{( 2 g D)}

where, f = Darcy friction factor

V = flow velocity

g = acceleration due to gravity

We know, flow rate Q = A x V

solving for V

V = \frac{Q}{A}

    = \frac{0.16}{\frac{\pi}{4} (0.318)^2} = 2.015 m/s

obtained Darcy friction factor  

calculate Reynold number (Re) ,

Re = \frac{\rho V D}{\mu}

where,\rho = density of water

\mu = Dynamic viscosity of water at 15 degree  C = 0.001 Ns/m2

so reynold number is

Re = \frac{1000\times 2.015\times 0.318}{0.001}

            = 6.4 x 10^5

For Schedule 80 PVC pipes , roughness (e) is  0.0015 mm

Relative roughness (e/D) = 0.0015 / 318 = 0.00005

from Moody diagram, for Re = 640000 and e/D = 0.00005 , Darcy friction factor , f = 0.0126

Therefore head loss is

HL = \frac{0.0126 (200)(2.015)^2}{( 2 \times 9.81 \times 0.318)}

HL = 1.64 m

7 0
4 years ago
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