# Make the internet a better place to learn

1

## What are seismoreceptors?

Sarthak
Featured 7 hours ago · Biology

siesmoreceptors can detect vibrational activity

#### Explanation:

Many animals communicate by producing Rayleigh waves (type of siesmic waves) on the substrate. This substrate can be anything including earth, spiders web, water surface, etc. Also it can be a mechanism of food detection as in spiders. They detect prey with the help of vibrations in the web created when an insect gets caught.

Elephants can detect seismic signals very well. They produce sounds with such high amplitudes that these can travel through the earth. These can be received by other elephants and as elephants have the largest cerebral cortex among all mammals, they can interpret these signals.

Many other animals also use seismic cues for different purposes. For more information, refer to this link.

1

## How much work would it take to horizontally accelerate an object with a mass of 8 kg to 3 m/s on a surface with a kinetic friction coefficient of 6 ?

Nathan L.
Featured 19 hours ago · Physics

$W = 36$ $\text{J}$ $+ 471 \textcolor{w h i t e}{l} \text{N" xx "distance}$

#### Explanation:

We're asked to find the necessary work to accelerate an $8$-$\text{kg}$ object from $0$ to $3$ $\text{m/s}$ on a surface with a coefficient of kinetic friction of $6$.

The equation for work according to the work-energy theorem is given by

ul(W = DeltaE_k = 1/2m(v_2)^2 - 1/2m(v_1)^2

The final velocity is $3$ $\text{m/s}$, and its mass is $8$ $\text{kg}$, so we have

W = 1/2(8color(white)(l)"kg")(3color(white)(l)"m/s" - 0)^2 = color(red)(ul(36color(white)(l)"J"

However, this does not include the work done to counteract the friction, so we have to use the equation

ul(W = Fs

The necessary force $F$ is

$F = {f}_{k} = {\mu}_{k} m g = \left(6\right) \left(8 \textcolor{w h i t e}{l} {\text{kg")(9.81color(white)(l)"m/s}}^{2}\right) = 471$ $\text{N}$

And so we have

$W = \left(471 \textcolor{w h i t e}{l} \text{N}\right) s$

The distance it travels $s$ can be any value because the applied force can be any number greater than the necessary force. Thus, we can leave it in terms of $s$:

W_"necessary" = color(blue)(ulbar(|stackrel(" ")(" "36color(white)(l)"J" + (471color(white)(l)"N")s" ")|)

1

## What are the things must be remember while studying adiabatic processes?

Truong-Son N.
Featured 19 hours ago · Physics

Well, it's always important to remember the definition of an adiabatic process:

$q = 0$,

So, there is no heat flow in or out (the system is thermally insulated from the surroundings).

From the first law of thermodynamics:

$\Delta E = q + w = q - \int P \mathrm{dV}$

where $w$ is the work from the perspective of the system and $\Delta E$ is the change in internal energy.

For an adiabatic process, we then have $\underline{\Delta E = w}$, so if the system expands, the internal energy of the system decreases as a direct result of the expansion work only.

From the second law of thermodynamics:

$\Delta S \ge \frac{q}{T}$

where $>$ corresponds to irreversible processes and $=$ corresponds to reversible processes.

If there is absolutely no heat flow in or out, the system entropy should be constant, as long as the process is reversible. You can read more about reversible processes here.

2

## Do all the enzyme cofactors get attached to the allosteric site of an enzyme?

Kasuni D.
Featured 7 hours ago · Biology

No

#### Explanation:

In the enzyme reactions many enzymes want a help from another group that isn't a protein.(non-proteins)
We call this group as the co-factor. It helps the enzyme in the enzyme reaction.

There are three main groups of co-factors.

1)Inorganic Ions
2)CoEnzymes
3)Prosthetic groups

Inorganic ions

This groups act as enzyme activators. Inorganic ions should bind to the allosteric site to activate the enzyme.
Many of inorganic ions are metal ions.

Ex:-
$Z {n}^{+}$
$F {e}^{2 +}$
$F {e}^{3 +}$

Co-Enzymes

These groups do not attach with the enzyme. These are organic molecules.
Co-enzymes can act with many enzymes. It means it is not specific to an enzyme.

If it released ${H}^{+}$ ions and electrons in a reaction with a enzyme,
then it changes into a different form.
And then it can help to another enzyme.
In that reaction the co-enzyme takes ${H}^{+}$ ions and electrons again and changes into the original form.

* *

Prosthetic groups

Prosthetic groups are non-proteins. But they may be inorgonic or orgonic groups.

They tightly bind to the enzyme and We can see a similar process as the above group, in the process of helping the enzyme by prosthetic groups.

But it does not get attached to the allosteric site. Normally it binds to the active site of the enzyme.

Ex:- an enzyme with a $h e m e$ group.

1

## When you study all the growth characteristics of a human population, what are you studying?

G_Ozdilek
Featured 7 hours ago · Environmental Science

Population dynamics

#### Explanation:

Population studies cover many issues, such as total population, population growth, population density, average (median) age, aged vs. young people, doubling time, crude birth rate, total fertility rate, crude death rate, age structure, etc.

The starting point in a population estimate (projection) is the current (today's) age structure and mortality data which could be taken fromlife tables. A life table is developed by applying a real population's age-specific death rates to hypothetical stable and stationary populations having 100,000 live births per year, evenly distributed through the year, with no migration. As the 100,000 people added each year aged, their ranks are thinned in accordance with the age-specific death rates.

Population studies provide details necessary for city planning, water supply, waste need to be collected, numbers of dwellings, roads, parking space, green areas (public parks), infrastructure, etc. By the way today's total (human) population is nearly 7,416,100,000 and we consume the resources and pollute the environment badly.

1

## Two wires X and Y have the same resistivity but their cross-sectional area are in the ratio 2:3 and length in the ratio 1:2. They are connected in series to a D.C. source . Find the ratio of their drift speeds of the eletrons in the two wires?

Steve J.
Featured 19 hours ago · Physics

The ratio of the drift speeds is $\frac{9}{4}$, the electrons in the narrow wire being faster.

#### Explanation:

Imagine a counter at both point A in the narrower wire and point B in the thicker wire. The counters are detecting and counting each electron as they pass in a 1 second period. The totals recorded by the 2 counters at the end of that second must be equal because the wires are in series, so the current thru both must be equal. And current is measured in Amps, which is equivalent to Coulombs/second.

Visualize the wire upstream from counters A and B just before the counters are started. The electrons that will be counted are in a cylinder in their individual wires. The formula for the volume of a cylinder is
$V = \text{cross section area"*"Length}$
Call those cylinders A and B and use the subscripts A and B to designate their volume, radii, and lengths. For comparison purposes, let ${r}_{A} = 2 \cdot r$ and ${r}_{B} = 3 \cdot r$.

The volumes of them are
${V}_{A} = \left(\pi \cdot {\left(2 \cdot r\right)}^{2}\right) \cdot {L}_{A}$
and
${V}_{B} = \left(\pi \cdot {\left(3 \cdot r\right)}^{2}\right) \cdot {L}_{B}$
The 2 electrons that will be the last electron counted by counter A and B were a distance ${L}_{A} \mathmr{and} {L}_{B}$ away, respectively, from their counter at the beginning of the 1 second period.

We assume that the same metal is used for the 2 wires. That assumption is required for another fact that we need to use: The density of free electrons in the 2 cylinders is equal because both use the same metal (most likely copper). That fact, with the fact that the number of electrons counted by the 2 counters is equal, give us the ability to say that the volume of the 2 cylinders are equal.

Therefore $\left(\pi \cdot {\left(2 \cdot r\right)}^{2}\right) \cdot {L}_{A} = \left(\pi \cdot {\left(3 \cdot r\right)}^{2}\right) \cdot {L}_{B}$
Cancelling where possible
$\frac{\cancel{\pi} \cdot {\left(2 \cdot \cancel{r}\right)}^{2}}{L} _ A = \frac{\cancel{\pi} \cdot {\left(3 \cdot \cancel{r}\right)}^{2}}{L} _ B$
yielding ${L}_{A} / {L}_{B} = \frac{9}{4}$

Since in each cylinder, the 2 electrons that were last to be counted in their section of wire traveled the distance ${L}_{A} \mathmr{and} {L}_{B}$ respectively, in 1 second, the ratio of the drift speeds is $\frac{9}{4}$.

I hope this helps,
Steve

2

## Suppose a quadratic function f(x)=x^2+bx+c has zeros p and q? let k be an integer. using only b,c,k, and x, write a new quadratic function who's zeros are p+k and q+k

Ratnaker Mehta
Featured yesterday · Algebra

${x}^{2} + \left(b - 2 k\right) x + \left({k}^{2} - b k + c\right) .$

#### Explanation:

We know that, if $p \mathmr{and} q$ are the Zeroes of the Quadr. Fun.

$f \left(x\right) = {x}^{2} + b x + c ,$ then,

$p + q = - \frac{b}{1} = - b , \mathmr{and} , p q = \frac{c}{1} = c \ldots \ldots \ldots \ldots \ldots \ldots . . \left(\ast\right) .$

Let, $\alpha = p + k , \mathmr{and} , \beta = q + k , \text{ where, } k \in \mathbb{Z} .$

Now, using $\left(\ast\right) ,$ we find,

$\therefore \alpha + \beta = \left(p + k\right) + \left(q + k\right) = p + q + 2 k = - b + 2 k \ldots \left({\ast}^{1}\right) ,$

and,

$\alpha \cdot \beta = \left(p + k\right) \left(q + k\right) = {k}^{2} + \left(p + q\right) k + p q = {k}^{2} - b k + c \ldots \left({\ast}^{2}\right) .$

Recall that, the Quadr. Fun., say, $g \left(x\right) ,$ whose Zeroes are

$\alpha , \mathmr{and} , \beta ,$ is given by,

$g \left(x\right) = {x}^{2} - \left(\alpha + \beta\right) x + \left(\alpha \cdot \beta\right) .$

Utilising, $\left({\ast}^{1}\right) , \mathmr{and} , \left({\ast}^{2}\right) ,$ we get, the desired fun.,

$g \left(x\right) = {x}^{2} - \left(- b + 2 k\right) x + \left({k}^{2} - b k + c\right) , i . e . ,$

$g \left(x\right) = {x}^{2} + \left(b - 2 k\right) x + \left({k}^{2} - b k + c\right) .$

Enjoy Maths.!

2

## What are continued fractions for?

Ken C.
Featured yesterday · Calculus

As far as I'm concerned, they're not really that important. But then again, I have no exposure to graduate-level concepts, and they may be useful there.

#### Explanation:

Nevertheless, for math nerds like me, they present some very interesting problems and manifest some very beautiful patterns. For instance, the continued (infinite) fraction:
1+1/(1+1/(1+...)

Is very famous because it simplifies to $\phi$, or the Golden Ratio ($1.618 \ldots$). To see why this is, we will employ a very clever trick. We begin by setting the fraction equal to $x$, as we would do with any unknown entity:
x=1+1/(1+1/(1+...)

But check this out. This part in red:
$1 + \frac{1}{\textcolor{red}{\left(1 + \frac{1}{1 + \ldots}\right)}}$

Is really just equal to $x$! Let's take it out of context and add a few more terms to see why:
color(red)(1+1/(1+1/(1+1/(1+...))

See? It's just the same infinite repeating fraction, so we replace it with $x$:
x=1+1/color(red)((1+1/(1+...))
$\to x = 1 + \frac{1}{\textcolor{red}{x}}$

Now we solve, through these steps:
${x}^{2} = x + 1$
$\to {x}^{2} - x - 1 = 0$
$\to x = \frac{{\left(- 1\right)}^{2} \pm \sqrt{{\left(- 1\right)}^{2} - 4 \left(1\right) \left(- 1\right)}}{\left(2\right) \left(1\right)}$
$\to x = \frac{1 \pm \sqrt{5}}{2}$

The $\frac{1 - \sqrt{5}}{2}$ solution is a negative number, and obviously the infinite fraction is not negative; therefore, our solution is
$\frac{1 + \sqrt{5}}{2}$
which is the number $\phi$ in radical form.

In fact, you can also express such math celebrities as $\pi$ and $e$ as infinite fractions, and discover some cool patterns like the Fibonacci sequence in them too. As for continued fractions in general, you can use them to approximate mathematical constants but you would be punishing yourself when other techniques (e.g. Taylor series-derived expansions) are way easier. I'm not really sure how else they're useful; infinite fractions are, to the extent of my knowledge, the most useful type.

1

## How do you graph y>=1/5x+10 on the coordinate plane?

smendyka
Featured 9 hours ago · Algebra

See a solution process below:

#### Explanation:

First, solve for two points as an equation instead of a inequality to find the boundary line for the inequality.

For $x = 0$

$y = \left(\frac{1}{5} \cdot 0\right) + 10 = 0 + 10 = 10$ or $\left(0 , 10\right)$

For $x = - 10$

$y = \left(\frac{1}{5} \cdot - 10\right) + 10 = - 2 + 10 = 8$ or $\left(- 10 , 8\right)$

We can now graph the two points on the coordinate plane and draw a line through the points to mark the boundary of the inequality.

The boundary line will be solid because the inequality operator contains a "or equal to" clause.

graph{(y-1/5x-10)((x+10)^2+(y-8)^2-0.3)(x^2+(y-10)^2-0.3)=0 [-30, 30, -15, 15]}

To complete the chart of the inequality we shade the left side of the line:

graph{(y-1/5x-10)>=0 [-30, 30, -15, 15]}

1

## N2O4=2NO2 A reaction mixture at 2000 degree C initially contains [NO2]=0.1 M. Find the equilibrium concentration of the reactant and products at this temperature? Kc =0.36 at 2000 degree C.

Nathan L.
Featured 21 hours ago · Chemistry

${\text{N"_2"O}}_{4} : 0.33$ $M$

${\text{NO}}_{2} : 0.34$ $M$

#### Explanation:

We're asked to find the equilibrium concentrations of ${\text{N"_2"O}}_{4}$ and ${\text{NO}}_{2}$, given the initial ${\text{NO}}_{2}$ concentration.

The equilibrium constant expression is given by

K_c = (["NO"_2]^2)/(["N"_2"O"_4]) = ul(0.36)" " $\left(2000 \textcolor{w h i t e}{l} \text{^"o""C}\right)$

We'll do our I.C.E. chart in the form of bullet points, for fun. Then, our initial concentrations are

INITIAL

• ${\text{N"_2"O}}_{4} :$ $0$

• ${\text{NO}}_{2} :$ $1$ $M$

According to the coefficients of the reaction, the amount by which ${\text{NO}}_{2}$ decreases is two times as much as the amount by which ${\text{N"_2"O}}_{4}$ increases:

CHANGE

• ${\text{N"_2"O}}_{4} :$ $+ x$

• ${\text{NO}}_{2} :$ $- 2 x$

And so the final concentrations are

FINAL

• ${\text{N"_2"O}}_{4} :$ $x$

• ${\text{NO}}_{2} :$ $1$ $M$ $- 2 x$

Plugging these into the equilibrium constant expression gives us

${K}_{c} = \frac{{\left(1 - 2 x\right)}^{2}}{x} = \underline{0.36} \text{ }$ (2000color(white)(l)""^"o""C", excluding units)

Now we solve for $x$:

$\frac{4 {x}^{2} - 4 x + 1}{x} = 0.36$

$4 {x}^{2} - 4 x + 1 = 0.36 x$

$4 {x}^{2} - 4.36 x + 1 = 0$

$x = \frac{4.36 \pm \sqrt{{\left(4.36\right)}^{2} - 4 \left(4\right) \left(1\right)}}{8} = 0.328 \textcolor{w h i t e}{l} \text{or} \textcolor{w h i t e}{l} 0.762$
If we plug the larger solution in for $x$ in the final ${\text{NO}}_{2}$ concentration $\left(1 - 2 x\right)$, we would obtain a worthless negative value. Therefore, we use the smaller solution to calculate the final concentrations:
color(red)("final N"_2"O"_4) = x = color(red)(ulbar(|stackrel(" ")(" "0.33color(white)(l)M" ")|)
color(blue)("final NO"_2) = 1-2x = 1-2(0.328) = color(blue)(ulbar(|stackrel(" ")(" "0.34color(white)(l)M" ")|)
each rounded to $2$ significant figures.