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3 years ago

What Is Photosynthesis ?

Engineering

2 answers:

Fynjy0 [20]3 years ago

6 0

Answer:

Photosynthesis is a process used by plants and other organisms to convert light energy into chemical energy that, through cellular respiration, can later be released to fuel the organism's activities.

Juliette [100K]3 years ago

4 0

Answer:

Photosynthesis is the process by which plants use sunlight, water, and carbon dioxide to create oxygen and energy in the form of sugar.

have a nice day <3

You might be interested in

What is programming and sensory?

NNADVOKAT [17]

Answer:

Programing is using data or codes to make something happen A sensory is like when you move your arm in front of a device and like it has a light that turns on then that is sensory it like can tell your moving

7 0

3 years ago

The equation for the velocity V in a pipe with diameter d and length L, under laminar condition is given by the equation V=Δpdsq

Citrus2011 [14]

Answer:

Given that

$V=\dfrac{\Delta Pd^2}{32\mu L}$

LHS of above given equation have dimension $[M^oL^{1}T^{-1}]$ .

Now find the dimension of RHS

Dimension of P = $[ML^{-1}T^{-2}]$ .

Dimension of d= $[M^{0}L^{1}T^{0}]$ .

Dimension of μ = $[ML^{-1}T^{-1}]$ .

Dimension of L= $[M^{0}L^{1}T^{0}]$ .

$\dfrac{\Delta Pd^2}{32\mu L}=\dfrac{[ML^{-1}T^{-2}].[M^{0}L^{1}T^{0}]^2}{[ML^{-1}T^{-1}].[M^{0}L^{1}T^{0}]}$

$\dfrac{\Delta Pd^2}{32\mu L}=[M^0L^{1}T^{-1}]$

It means that both sides have same dimensions.

5 0

4 years ago

Multiply each element in origList with the corresponding value in offsetAmount. Print each product followed by a space.Ex: If or

n200080 [17]

Answer:

Here is the JAVA program:

import java.util.Scanner; // to take input from user

public class VectorElementOperations {

public static void main(String[] args) {

final int NUM_VALS = 4; // size is fixed that is 4 and assigned to NUM_VALS

int[] origList = new int[NUM_VALS];

int[] offsetAmount = new int[NUM_VALS];

int i;

//two lists origList[] and offsetAmount[] are assigned values

origList[0] = 20;

origList[1] = 30;

origList[2] = 40;

origList[3] = 50;

offsetAmount[0] = 4;

offsetAmount[1] = 6;

offsetAmount[2] = 2;

offsetAmount[3] = 8;

String product=""; // product variable to store the product of 2 lists

for(i = 0; i <= origList.length - 1; i++){

/* loop starts with i at 0th position or index and ends when the end of the origList is reached */

/* multiples each element of origList to corresponding element of offsetAmount and stores result in the form of character string in product*/

product+= Integer.toString(origList[i] *= offsetAmount[i]) + " "; }

System.out.println(product); }} //displays the product of both lists

Output:

80 180 80 400

Explanation:

If you want to print the product of origList with corresponding value in offsetAmount in vertical form you can do this in the following way:

import java.util.Scanner;

public class VectorElementOperations {

public static void main(String[] args) {

final int NUM_VALS = 4;

int[] origList = new int[NUM_VALS];

int[] offsetAmount = new int[NUM_VALS];

int i;

origList[0] = 20;

origList[1] = 30;

origList[2] = 40;

origList[3] = 50;

offsetAmount[0] = 4;

offsetAmount[1] = 6;

offsetAmount[2] = 2;

offsetAmount[3] = 8;

for(i = 0; i <= origList.length - 1; i++){

origList[i] *= offsetAmount[i];

System.out.println(origList[i]); } }}

Output:

180

400

The program along with the output is attached as screenshot with the input given in the example.

8 0

3 years ago

The beam is supported by a pin at A and a roller at B which has negligible weight and a radius of 15 mm. If the coefficient of s

Anettt [7]

Answer:

33.4

Explanation:

Step 1:

\sumMo=0 (moment about the origin)

Fb(15)-Fc(15)=0

Fb=Fc

Step 2:

\sumFx=0

-Fb-Fccos\theta+Ncsin\theta=0

Fc=0.3Nc=Fb

-0.3Nc-0.3Nccos\theta+Ncsin\theta=0

(-0.3-cos\theta+sin\theta)Nc=0----(1)

\sumFy=0

Nccos\theta+Fcsin\theta-Nb=0

Nccos\theta+0.3Ncsin\theta-Nc=0

Nc[cos\theta+0.3sin\theta-1]=0--------(2)

Solving eq (1) and eq (2)

\theta=33.4

Step 3:

As the roller is a two force member

2(90-\phi)+\theta=180

\phi=\theta/2

\phi=Tan(\muN/N)-1

\phi=16.7

\theta=2x16.7=33.4

5 0

3 years ago

A heat engine uses fuel of energy content 43.1 MJ/kg and produces 17.4 kW of useful power. The heat rejection rate (through the

solmaris [256]

Answer:

a)27.9%

b) $\dot{m}=3.06 \frac{Kg}{h}$

Explanation:

Given that

Fuel energy content = 73.1 MJ/kg

Useful power = 17.4 KW

Heat rejection rate = 44.8 KW

From first law of thermodynamics

Heat addition rate =Heat rejection rate + Power out put

Now by putting the values in the above formula

Heat addition rate = 44.8 + 17.4

Heat addition rate =62.2 KW

We know that efficiency is given as follows

$\eta =\dfrac{power\ out\ put}{Heat\ addition\ rate}$

$\eta =\dfrac{17.4}{62.2}$

$\eta =0.279$

So the efficiency is 27.9%.

Now to find usage rate of fuel

Lets take usage rate is $\dot{m}$

Fuel energy content x usage rate of fuel = Heat addition rate

Now by putting the values

$73100\times \dfrac{\dot{m}}{3600}=62.2$

$\dot{m}=3.06 \frac{Kg}{h}$

7 0

3 years ago

What Is Photosynthesis ?​

What Is Photosynthesis ?