32

## Why is the water cycle important to all life on the earth?

Kate M.
Featured 8 months ago · Environmental Science

The water cycle is important to all life on earth for many reasons.

#### Explanation:

The water cycle is important to all life on earth for many reasons. All living organisms require water and the water cycle describes the process of how water moves through the planet.

Plants wouldn't grow without precipitation (and thus anything consuming the plants wouldn't survive and so forth). Infiltration of water filters and cleans our water. Glaciers, ice, and snow can act as stores of freshwater for both humans and other organisms. Runoff contributes to rivers, other freshwater bodies, and eventually the ocean, sustaining freshwater and marine life.

All of these process sustain life and create the ecosystems around us. Some organisms are very sensitive to changes in the water cycle. A prolonged drought can destroy a population of plants or a certain salamander species may require a set amount of soil saturation in order to avoid desiccation.

See this question on how humans use water in everyday life, this webpage from NASA on the water cycle, and this PDF on how the natural world filters water for more information on why the water cycle is so important.

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1

## What happens during the Calvin-Benson cycle, or dark reaction?

Arsala K.
Featured 2 days ago · Biology

In Calvin cycle or dark reaction, sugars are formed by using the chemical energy of $A T P$ and $N A D P H$ formed during light reaction.

#### Explanation:

Equation for dark reaction:

$3 C {O}_{2}$ + $6 N A D P H$ + $9 A T P \to {\left(C {H}_{2} O\right)}_{3}$ + $6 N A D P H$ +
$9 A D P$ +$9$Pi

Diagram of Calvin cycle:

Steps in Calvin Cycle:

1st step: Carbon fixation:

It refers to the initial incorporation of $C {O}_{2}$ into organic material. We will follow $3$ molecules of $C {O}_{2}$ through the reaction because we want sugars(carbohydrates) as the end product of cycle. And for this, at least 3 molecules of $C {O}_{2}$ are necessary.
Because;
$3$ molecules of $C {O}_{2}$→ contain $3$ C→ so at least they will give $1$ carbohydrate molecule( a triose).

• Calvin cycle begins by the reaction of CO2 with highly reactive phosphorylated five carbon-sugar named ribulose bisphosphate(RuBP).

• This reaction is catalyzed by the enzyme ribulose bisphosphate carboxylase also known as Rubisco .

• The product of this reaction is highly unstable six-carbon intermediate that immediately breaks into two molecules of three-carbon-compound named 3-phosphoglycerate(PGA).
• The carbon that was originally a part of $C {O}_{2}$ molecule is now a part of an organic compound which means carbon has been fixed .

2nd step: Reduction:

In this step, fixed carbon is reduced to energy rich G3P with the energy and reducing power of ATP and NADPH respectively.

• Each molecule of PGA recieves an additional phosphate from ATP of light reaction and form 1,3-bisphosphoglycerate.

• It is reduced to glyceraldehyde 3-phosphate(G3P) by recieving a pair of electrons donated from NADPH of light reactions.
Actually G3P is the carbohydrate that is produced directly from Calvin cycle.

• For every $3$ molecules of $C {O}_{2}$ entering the cycle and combining with $3$ molecules of RuBp, $6$ molecules of G3P are produced. But only $1$ molecule of G3P can be counted as a net gain of carbohydrate.
Because out of every $6$ molecules of G3P formed, only one molecule leaves the cycle to be used by plants for making glucose and other carbohydrates; the other five molecules are recycled to regenerate the three molecules of five carbon RuBP, the $C {O}_{2}$ acceptor.

3rd step: Regeneration of $C {O}_{2}$ acceptor, RuBp:

• Through a complex series of reactions, the carbon skeletons of five molecules of three-carbon G3P are rearranged into three molecules of five-carbon ribulose phosphate(RuP).

• Each RuP is phosphorylated to RuBP. Again $3$ molecules of ATP of light reactions are used for this phosphorylation of $3$
RuP.

• These RuBP are now prepared to recieve $C {O}_{2}$ again, and the cycle continues.

1

## How can recessive, lethal diseases continue to exist in a population?

BillytheKid
Featured 2 days ago · Biology

Because we have a double set of Chromosomes.

#### Explanation:

If a gene is recessive it usually is broken/inactive/damaged.
As long as you have a dominant, proper working gene on the corresponding chromosome donated by your other parent, you will survive and pass the recessive gene on to the next generation...

To review the basics of inheritance, see here and Genetics Overview.

Though not necessarily lethal anymore these days, Haemophilia is a relatively good example: This is caused by a defective gene that normally codes for FactorVIII (8), an essential factor in bloodclotting.

It resides on the X-chromosome....

Females have 2 X-chromosomes, so if one carries a defective gene, it will be compensated for by the other chromosome.

Males aren't so lucky: They receive an X-chromosome from their mother and a Y-chromosome from their father.

The Y-chromosome doesn't contain the FactorVIII gene, so if you inherit a defective one from your mother ( on the X-chromosome she contributed), you're in trouble....

Females who have a defective gene on one of their X-chromosomes won't be Haemophiliac, but they are a Carrier.

Occasionally a female CAN inherit TWO X-chromosomes that have defective FVIII-genes, and then she will be affected, but the chance is much lower...

As mentioned, in this case the gene in question resides on a sex-linked chromosome, but same goes up for Autosomal Chromosomes. In that case both males and females will be equally affected....

1

## What were 4 resources the North had more of?

David Drayer
Featured 11 hours ago · U.S. History

men, medicine, rifles, ships.

#### Explanation:

1. Grant could afford to lose battle after battle to Lee in his drive to Petersburg, because Grant could replace his loses, while the south could not Because of immigration the north had much larger supplies of replacement soldiers.

2. Because of the anaconda plan the south was blocked from imports. The south was not able to get medical supplies like opium and other medical supplies. Many soldiers died or loss limbs in the south because of the lack of medical supplies. Sadly many soldiers in the north became addicted to opium. The over supply of medical supplies in the north had some negative consequences.

3. rifles, cannons, were produced in factories in the north. Also the north could import arms from Europe. The south had a few factories that could produce weapons. The blockade meant that it was difficult for the south to import arms from Europe. After the fall of Vicksburg the south could not get any further arms from Europe. The soldiers in the north had access to repeating rifles, better muskets, Parrot cannons, colt revolvers, and sharp shooting rifles with scopes. All of the weapons that the northern army used were superior to the weapons available to the soldiers in the southern armies.

4. The armored gun boats of the Union dominated the Mississippi River conflict. The blockade was possible because the Union had many more ships than were available to the south.

1

## The ideal gas law states pv=nrt, find how fast p is changing when the volume is 10 m^3, p=20 pascals, and volume is decreasing at a rate of 2 m^3 per min, and temperature is increasing at a rate of 4 kelvin per min. How do I solve this?

Jason K.
Featured 3 days ago · Calculus

$\frac{\mathrm{dP}}{\mathrm{dt}} = \frac{2}{5} n R + 4$

#### Explanation:

In the Ideal Gas Law formula, there are 3 variables and 2 constants. $P$ represents the pressure (in this problem in units of pascals), $V$ represents the volume (in this problem in units of ${m}^{3}$), and $T$ represents the temperature (in this problem in Kelvin). We consider $n$ to be the amount of gas present (usually in moles), while $R$ is the Ideal Gas Constant, or a fixed value.

This is a "classic" related rates problem. We need to take derivatives of every variable ($P , V , T$) with respect to time $t$. To do so, we perform "traditional" derivatives on each of those terms in the equation, but we ensure that we include a $\frac{\mathrm{dP}}{\mathrm{dt}}$, $\frac{\mathrm{dV}}{\mathrm{dt}}$, or $\frac{\mathrm{dT}}{\mathrm{dt}}$ term as necessary.

$P V = n R T$

$P {\underbrace{\left(1 \cdot \frac{\mathrm{dV}}{\mathrm{dt}}\right)}}_{\text{d(V)" + Vunderbrace((1*(dP)/(dt)))_ "d(P)" = nRunderbrace((1*(dT)/(dt)))_ "d(T)}}$

$P \left(\frac{\mathrm{dV}}{\mathrm{dt}}\right) + V \left(\frac{\mathrm{dP}}{\mathrm{dt}}\right) = n R \left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)$

We have all of the values we need provided to us to substitute into this related rates expression:

$\left\{\begin{matrix}P & 20 \text{pascals" & "Pressure" \\ (dV)/(dt) & -2 m^3/min & "Volume Change" \\ V & 10 m^3 & "Volume" \\ (dP)/(dt) & ? & "Pres. Change" \\ (dT)/(dt) & 4 K/min & "Temp Change}\end{matrix}\right.$

$\left(20\right) \left(- 2\right) + \left(10\right) \left(\frac{\mathrm{dP}}{\mathrm{dt}}\right) = n R \left(4\right)$

$- 40 + 10 \left(\frac{\mathrm{dP}}{\mathrm{dt}}\right) = 4 n R$

$10 \left(\frac{\mathrm{dP}}{\mathrm{dt}}\right) = 4 n R + 40$

$\frac{\mathrm{dP}}{\mathrm{dt}} = \frac{4 n R + 40}{10} = \frac{2}{5} n R + 4$

A quick "sanity check" verifies that the sign of our answer is what we'd expect. Since P is inversely related to V and directly related to T, if the volume is decreasing, and the temperature is increasing, we'd definitely expect the pressure P to be increasing (positive). Since $n$ and $R$ are both positive constants, the final expression will be positive.

1

## How do i solve \int xln \frac{x+1}{1-x}dx?

Cem Sentin
Featured 3 days ago · Calculus

${x}^{2} / 2 \cdot L n \left[\frac{x + 1}{1 - x}\right] - {x}^{2} / 2 - \frac{1}{2} \cdot L n \left({x}^{2} - 1\right) + C$

#### Explanation:

$\int x \cdot \ln \left[\frac{x + 1}{1 - x}\right] \cdot \mathrm{dx}$

After setting $\mathrm{dv} = x \cdot \mathrm{dx}$ an $u = L n \left[\frac{x + 1}{1 - x}\right] = L n \left(x + 1\right) - L n \left(1 - x\right)$,

$v = {x}^{2} / 2$ and $\mathrm{du} = \frac{\mathrm{dx}}{x + 1} - \frac{- \mathrm{dx}}{1 - x} = \frac{\mathrm{dx}}{x + 1} + \frac{\mathrm{dx}}{x - 1} = \frac{2 x \cdot \mathrm{dx}}{{x}^{2} - 1}$

So,

$\int u \cdot \mathrm{dv} = u v - \int v \cdot \mathrm{du}$

$\int x \cdot \ln \left[\frac{x + 1}{1 - x}\right] \cdot \mathrm{dx} = {x}^{2} / 2 \cdot L n \left[\frac{x + 1}{1 - x}\right] - \int {x}^{2} / 2 \cdot \frac{2 x \cdot \mathrm{dx}}{{x}^{2} - 1}$

=${x}^{2} / 2 \cdot L n \left[\frac{x + 1}{1 - x}\right] - \int \frac{{x}^{3} \cdot \mathrm{dx}}{{x}^{2} - 1}$

=${x}^{2} / 2 \cdot L n \left[\frac{x + 1}{1 - x}\right] - \int \frac{\left({x}^{3} - x + x\right) \cdot \mathrm{dx}}{{x}^{2} - 1}$

=${x}^{2} / 2 \cdot L n \left[\frac{x + 1}{1 - x}\right] - \int \frac{\left({x}^{3} - x\right) \cdot \mathrm{dx}}{{x}^{2} - 1} - \int \frac{x \cdot \mathrm{dx}}{{x}^{2} - 1}$

=${x}^{2} / 2 \cdot L n \left[\frac{x + 1}{1 - x}\right] - \int x \cdot \mathrm{dx} - \frac{1}{2} \cdot \int \frac{2 x \cdot \mathrm{dx}}{{x}^{2} - 1}$

=${x}^{2} / 2 \cdot L n \left[\frac{x + 1}{1 - x}\right] - {x}^{2} / 2 - \frac{1}{2} \cdot L n \left({x}^{2} - 1\right) + C$

2

## What are Mexico's three major mountain ranges?

Dean
Featured yesterday · Earth Science

Mexico consists of an entire mountain system known as the Sierra Madre. The three ranges it consists of are called Sierra Madre Occidental, Sierra Madre Oriental, and Sierra Madre del Sur.

#### Explanation:

The mountain system is located in central Mexico. "Western" in Spanish is "occidental," meaning Sierra Madre Occidental is to the west. "Eastern" in Spanish is "oriental," meaning Sierra Madre Oriental is to the east. "Southern" in Spanish is "del sur," meaning Sierra Madre del Sur is to the south.

Fun Fact: Sierra Madre Oriental contains the largest point in Mexico called the Pico de Orizaba (or "Peak of Orizaba [Mexican city]").

1

## What is the correct order for -1\frac { 2} { 5} ,( - 1) ^ { 2} ,- 1.4 and ( \frac { 1} { 2}^2 )?

EZ as pi
Featured 2 days ago · Algebra

$- 1 \frac{2}{5} , \text{ "-1.4," "1^2/2," } 1$

#### Explanation:

Convert them into the same form. Decimals are easiest to compare.

$- 1 \frac{2}{5} , \text{ "(-1)^2," "-1.4," } \frac{{1}^{2}}{2}$

$= - 1.4 , \text{ "+1," "-1.4," } 0.5$

The two negative values are equal and are the smallest.

$- 1 \frac{2}{5} , \text{ "-1.4," "1^2/2," } 1$

1

## Ajay and Vijay have same amount of money with them. If Ajay gives ₹5 to Vijay,Ajay would have twice the amount that Vijay would have.If Vijay gives ₹15 to Ajay, Ajay would have eight times the amount that Vijay would have. Find the amount with ajay.?

Evan
Featured 2 days ago · Algebra

Ajay has ₹65.

#### Explanation:

Let Ajay's amount be $x$ and vijay's amount be $y$,

When Ajay gives Vijay ₹5,
$x - 5 = 2 \left(y + 5\right)$
$x - 5 = 2 y + 10$
color(white)(xxx)x=2y+15-①

When Vijay gives Ajay ₹15,
$x + 15 = 8 \left(y - 15\right)$
$x + 15 = 8 y - 120$
color(white)(xxx.)x=8y-135-②

①=②,
$2 y + 15 = 8 y - 135$
$\textcolor{w h i t e}{\times x .} 6 y = 150$
$\textcolor{w h i t e}{\times x \ldots} y = 25$

When $y = 25$,
$x = 2 \left(25\right) + 15$
$\textcolor{w h i t e}{x} = 65$

1

## Please how do I round 56.479 cents to the nearest cents?

Tony B
Featured 2 days ago · Prealgebra

56 cents

#### Explanation:

We have 56 cents + 0.479 cents

so the 'cut off' point is the red part

$\textcolor{red}{56} \textcolor{g r e e n}{.479}$

We look at the first digit only that is immediately to the right of the cut off point. In this case it is the four $\to \textcolor{g r e e n}{4}$

This is the rule:
If this digit is less than 5 we do not change the red part.
If the digit is 5 or more we increase the red part by 1.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{Example}}$

Making up a value I choose $\textcolor{red}{2} \textcolor{g r e e n}{.301}$
As the $\textcolor{g r e e n}{3}$ is less than 5 we do not change the red part. So rounded to the nearest whole number we have: $\textcolor{red}{2.0}$ to 0 decimal places

Making up a another value I choose $\textcolor{red}{2} \textcolor{g r e e n}{.601}$
As the $\textcolor{g r e e n}{6}$ is more than or equal to 5 we increase the red part by 1. So rounded to the nearest whole number we have: $\textcolor{red}{3.0}$ to 0 decimal places
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{Back to the question}}$

The $\textcolor{g r e e n}{4}$ from $\textcolor{red}{56} \textcolor{g r e e n}{.479}$
is less than 5 so we do not change the red part.

Answer $\to 56$

1

## Convert 50,000 seconds to days?

Jim G.
Featured 2 days ago · Prealgebra

$\approx 0.5787 \text{ days to 4 dec. places}$

#### Explanation:

$\text{using the following conversions}$

• " 1 minute "=60" seconds"

• " 1 hour "=60" minutes"

• " 1 day "=24" hours"

$\text{seconds "-:60to" minutes}$

$\text{minutes "-:60to" hours}$

$\text{hours "-:24to" days}$

$\text{thus dividing seconds by "(60xx60xx24)" gives days}$

$\Rightarrow \frac{50000}{60 \times 60 \times 24} = 0.5787037 \ldots .$

$\Rightarrow \approx 0.5787 \text{ days to 4 decimal places}$