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irga5000 [103]
2 years ago
6

The systems of the body work together and are dependent upon one another. Explain how another system, besides the skeletal syste

m, works with the muscular system.
Biology
1 answer:
Nimfa-mama [501]2 years ago
5 0

Active muscles demand huge amounts of oxygen, and the circulatory system works hard to provide it. Blood carries oxygen to the muscles. When the body is at rest, blood carries nutrients to the muscles so that they can repair and rebuild themselves. Blood also carries away the waste products that hard-working muscles produce, back out through the lungs to be filtered.

so they Circulatory system xx muscular system :)

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Leviafan [203]

Frogs have their way of avoiding the predators like camouflage, squirting the poisonous liquid out of their system and through their vestigial thumbs. This organ helps them by avoiding predation by stealth which makes them quiet and silent in moving so that predators won't notice them readily. 
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2 years ago
What happens to the oxygen and eagle ray takes in through its gills
Zanzabum

Answer:

Spotted Eagle Rays have two sets of five gills on each side of their body on the ventral side. In order to breathe, A. narinari must be in continuous swimming motion so that oxygen can be absorbed from the water passing through the gills

Explanation:

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2 years ago
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The average adult (like Hank) is _________% water, ________% protein, _______% fat, _______% minerals, and _______% carbohydrate
Oliga [24]

Answer: 64% water, 14% protein, 16% fat, 4% minerals 1% carbohydrates

Explanation:

It’s what he says in the video

3 0
1 year ago
Modify the existing vector's contents, by erasing the element at index 1 (initially 200), then inserting 100 and 102 in the show
Aleonysh [2.5K]

Answer:

Following are the code to this question:

#include <iostream>//defining header file

#include <vector>//defining header file

using namespace std;

void PrintVectors(vector<int> numsList)//defining a method PrintVectors that accept an array

{

   int j;//defining integer variable

   for (j = 0; j < numsList.size(); ++j)//defining for loop for print array value  

   {

       cout << numsList.at(j) << " ";//print array

   }

   cout << endl;

}

int main()//defining main method  

{

   vector<int> numsList;//defining array numsList

   numsList.push_back(101);//use push_back method to insert value in array

   numsList.push_back(200);//use push_back method to insert value in array

   numsList.push_back(103);//use push_back method to insert value in array

   numsList.erase(numsList.begin()+1);//use erase method to remove value from array

   numsList.insert(numsList.begin(), 100);//use insert method to add value in array

   numsList.insert(numsList.begin()+2, 102);//use insert method to add value in array

   PrintVectors(numsList);//use PrintVectors method print array value

   return 0;

}

Output:

100 101 102 103  

Explanation:

In the above-given code, inside the main method an integer array "numList" is defined, that use insert method to insert value and use the erase method to remove value from the array and at the last "PrintVectors" method is called that accepts a "numList" in its parameter. In the "PrintVectors" method, a for loop is declared, that prints the array values.

7 0
2 years ago
Pseudomonas sp. has a mass doubling time of 2.4 h when grown on acetate. The saturation constant using this substrate is 1.3 g/l
erastova [34]

Answer:

a)  Cell concentration when the dilution rate is one-half of the maximum is  0.598g cell/L

b) the substrate concentration when the dilution rate is 0.8 D_{max}   is 5.2g/l

c) the maximum dilution rate is : 0.41 h⁻¹

d)  the cell productivity at 0.8  D_{max}   is 2.40g cell/L

Explanation:

Given data :

Mass doubling time of Pseudomonas sp. = 2.4 hr

Saturation constant = 1.3 g/L

Cell yield  on acetate = 0.46g cell/g acetate

We are to find;

a. Cell concentration when the dilution rate is one-half of the maximum.

Here, cell yield =amount of cell produced / amount of substrate consumed.

[S] at 0.5D max is determined using the Monod's equation.

Using the formula :

D = \frac {D_{max}[S] }{ks+[S]}

, where D is the dilution rate,

[S] is substrate concentration; &

Ks is the saturation constant.

By replacing the values, we get :

0.5 = \frac{S}{1.3+[S]}

\\\0.65=0.5[S]

[S]=1.3g/L

The cell concentration at 0.5Dmax= cell yield x substrate consumed at 0.5Dmax.

=0.46×1.3

= 0.598g cell/L

b)

Substrate concentration when the dilution rate is 0.8 D_{max}    is calculated as:

D = \frac {D_{max}[S] }{ks+[S]}

0.8=[S]/1.3+[S]

1.04+0.8[S]=[S]

[S]= 5.2g/L

Therefore ,  the substrate concentration when the dilution rate is 0.8 D_{max}   is 5.2g/l

c)

Maximum dilution rate is calculated using the expression D_{max} = \frac{1}{time}

=1/2.4

=0.41 h⁻¹

So, the maximum dilution rate is : 0.41 h⁻¹

d)

The cell productivity at 0.8 D_{max} can be calculated by multiplying the amount  of the cell yield with the amount of the substrate consumed at 0.8D_{max}  

Cell yield = \frac {cell \ productivity \ at \  0.8Dmax}{amount \ of \ substrate\ consumed \at\ 0.8 \D_{max}}

Cell productivity at 0.8 D_{max}    = 0.46 × 5.2

=2.40g cell/L

Therefore, the cell productivity at 0.8  D_{max}   is 2.40g cell/L

5 0
2 years ago
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