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jarptica [38.1K]

2 years ago

8

About what fraction of earth's freshwater is in the form of ice

1 answer:

Bingel [31]2 years ago

3 0

68% So 68/100 of freshwater is found in ice

You might be interested in

A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i

Arturiano [62]

Answer:

$q / q_{1} = 2$ , assuming that $q_{1}$ and $(q - q_{1})$ are point charges.

Explanation:

Let $k$ denote the coulomb constant. Let $r$ denote the distance between the two point charges. In this question, neither $k$ and $r$ depend on the value of $q_{1}$ .

By Coulomb's Law, the magnitude of electrostatic force between $q_{1}$ and $(q - q_{1})$ would be:

$\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}$ .

Find the first and second derivative of $F$ with respect to $q_{1}$ . (Note that $0 < q_{1} < q$ .)

First derivative:

$\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}$ .

Second derivative:

$\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}$ .

The value of the coulomb constant $k$ is greater than $0$ . Thus, the value of the second derivative of $F$ with respect to $q_{1}$ would be negative for all real $r$ . $F\!$ would be convex over all $q_{1}$ .

By the convexity of $\! F$ with respect to $\! q_{1} \!$ , there would be a unique $q_{1}$ that globally maximizes $F$ . The first derivative of $F\!$ with respect to $q_{1}\!$ should be $0$ for that particular $\! q_{1}$ . In other words:

$\displaystyle \frac{k}{r^{2}}\, (q - 2\, q_{1}) = 0$ <em>.</em>

$2\, q_{1} = q$ .

$q_{1} = q / 2$ .

In other words, the force between the two point charges would be maximized when the charge is evenly split:

$\begin{aligned} \frac{q}{q_{1}} &= \frac{q}{q / 2} = 2\end{aligned}$ .

3 0

2 years ago

Encuentra la masa molar<br><br> TO<br><br> de 65 litros de SO2

shepuryov [24]

La masa molar de 65 litros de SO2 es igual a 64,1 g/mol.

<h3>Masa molar</h3>

La masa molar de un compuesto depende de su masa presente en 1 mol, entonces:

$MM=\frac{m}{mol}$

Para calcular la masa molar de un compuesto, simplemente suma las masas de cada elemento en el compuesto, así:

$MM_O = 16g/mol\\MM_S= 32.1g/mol$

$32.1 + 2\times 16 = 64.1g/mol$

Así, la masa molar de 65 litros de SO2 es igual a 64,1 g/mol.

Obtenga más información sobre la masa molar en: brainly.com/question/17109809

8 0

2 years ago

A thin film of cryolite ( nc = 1.34 ) is applied to a camera lens ( ng = 1.58 ). The coating is designed to reflect wavelengths

rosijanka [135]

Answer:

=99.07nm

Explanation:

minimum thickness

2nd = (m - 1/2)λ

d = (m - 1/2)(λ/2n)

refractive index of the thin film, n = 1.34

minimum thickness m = 1

light wavelength λ = 531nm

d = (1 - 1/2) (531 / (2)(1.34)

d = 531/5.36

= 99.07nm

3 0

2 years ago

An object moving at 30 m/s takes 5 sec. to come to a stop. What is the object’s acceleration

Klio2033 [76]

Answer:

6 m/s²

Explanation:

a=v/t

a= 30 m/s÷ 5 sec= 6 m/s²

4 0

3 years ago

A 35 kg box rests on the back of a truck. The coefficient of static friction bet?005 (part 1 of 2)A 35 kg box rests on the back

elena-14-01-66 [18.8K]

Answer with Explanation:

We are given that

Mass of box=35 kg

Coefficient of static friction between box and truck bed=0.202

Acceleration due to gravity= $9.8 m/s^2$

a.We have to find the force by which the box accelerates forward.

Force by which box accelerates= $\mu mg=0.202\times 9.8\times 35$

Force by which box accelerates= $62.286 N$

b.We have to find the maximum acceleration can the truck have before the box slides.

Force =friction force

$ma=\mu mg$

$a=\mu g=9.8\times 0.202=1.9796 m/s^2$

Hence, the truck can have maximum acceleration before the box slide= $1.9796 m/s^2$

3 0

3 years ago