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ivanzaharov [21]

3 years ago

14

This graph shows a supply curve. A graph titled Supply Curve has Quantity Supplied on the x-axis, from 0 to 50 in increments of

10, and price on the y-axis, from 0 to 15 dollars in increments of 5 dollars. A line is drawn with points (10, 5 dollars), (20, 7 dollars), (30, 10 dollars), (40, 13 dollars), and (40, 15 dollars). What happens when the price of a good increases? The quantity of goods that are produced increases. The producer of the good is certain to make less money. The quantity of goods that are produced decreases. The quantity of goods that are produced stays about the same.

1 answer:

Otrada [13]3 years ago

7 0

Answer:

A just took the quiz on edge

Explanation:

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Determine the normal boiling point of a substance whose vapor pressure is 55.1 mm hg at 35°c and has a δhvap of 32 .1 kj/mol. de

atroni [7]

Classius claperyon equation
In (P2/ P2) = ΔHvap/R) × (1/T2-1/T1)
T2 occurs at normal boiling when vapor pressure P2 = 1 atm.
P1 = 55.1 mmHg, P2 = 1 atm = 760mmHg
T1 = 35°c = 308.15k, T2 =
ΔHvap = 32.1kJ/mol = 32100 J/mol
In (760/55.1) = (-32100/ 8.314) × ( 1/T2 - 1/308.15)
The normal boiling point T2 = 390k = 117°c

5 0

3 years ago

Why do areas north of the artic circle in tje northern hemisphere experience a polar day lasting for several months during summe

adell [148]

<span>The regioin is titled towqrd the Sun during polar day. (C)

(The same exact thing happens in areas south of the Antarctic Circle
in the southern hemisphere. The only difference is that the whole thing
is spelled better in the South.)</span>

7 0

3 years ago

Read 2 more answers

Compare the circular velocity of a particle orbiting in the Encke Division, whose distance from Saturn 133,370 km, to a particle

Ket [755]

Answer:

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

Explanation:

The circular velocity is define as:

$v = \frac{2 \pi r}{T}$

Where r is the radius of the trajectory and T is the orbital period

To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:

$T^{2} = r^{3}$

Where T is orbital period and r is the radius of the trajectory.

Case for the particle in the Encke Division:

$T^{2} = r^{3}$

$T = \sqrt{(133370 Km)^{3}}$

$T = \sqrt{(2.372x10^{15} Km)}$

$T = 4.870x10^{7} Km$

It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( $1.50x10^{8} Km$ )

1 AU is defined as the distance between the earth and the sun.

$\frac{4.870x10^{7} Km}{1.50x10^{8}Km} . 1AU$

$T = 0.324 AU$

But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.

$T = \frac{0.324 AU}{1 AU} . 1 year$

$T = 0.324 year$

That can be expressed in units of days

$T = \frac{0.324 year}{1 year} . 365.25 days$

$T = 118.60 days$

<em>Circular velocity for the particle in the </em><em>Encke Division</em><em>:</em>

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (133370 Km)}{(118.60 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$118.60 days .\frac{86400 s}{1 day}$ ⇒ 10247040 s

$133370 Km .\frac{1000 m}{1 Km}$ ⇒ 133370000 m

$v = \frac{2 \pi (133370000 m)}{(10247040 s)}$

$v = 81.778 m/s$

Case for the particle in the D Ring:

For the case of the particle in the D Ring, the same approach used above can be followed

$T^{2} = r^{3}$

$T = \sqrt{(69000 Km)^{3}}$

$T = \sqrt{(3.285x10^{14} Km)}$

$T = 1.812x10^{7} Km$

$\frac{1.812x10^{7} Km}{1.50x10^{8}Km} . 1 AU$

$T = 0.120 AU$

$T = \frac{0.120 AU}{1 AU} . 1 year$

$T = 0.120 year$

$T = \frac{0.120 year}{1 year} . 365.25 days$

$T = 43.83 days$

<em>Circular velocity for the particle in </em><em>D Ring</em><em>:</em>

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (69000 Km)}{(43.83 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$43.83 days . \frac{86400 s}{1 day}$ ⇒ 3786912 s

$69000 Km . \frac{1000 m}{ 1 Km}$ ⇒ 69000000 m

$v = \frac{2 \pi (69000000 m)}{(3786912 s)}$

$v = 114.483 m/s$

$\frac{114.483 m/s}{81.778 m/s} = 1.399$

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

7 0

3 years ago

IF U ANSWER I WILL MARK U BRAINLLIEST

goldfiish [28.3K]

Explanation:

a) Density is the amount of mass in a specific amount of volume

b) Some substances are less dense than water so will float and vice versa

c) The wooden plank has less density than water due to having larger volume and vice versa with the wooden block since volume and density are inversely proportional

II) (a) They are all metals

(b) water -> Compound

oxygen -> Element

Gold-> Element

Salt-> Compound

6 0

3 years ago

Read 2 more answers

Que unidad de medida se utiliza para expresar la capacidad o volumen de un objeto? Da un ejemplo

galina1969 [7]

Answer:

<em>Explicado a continuación</em>

Explanation:

Hay una pequeña diferencia conceptual entre la capacidad y el volumen de un objeto, a saber:

El volumen hace referencia al espacio que ocupa un objeto, mientras que la capacidad hace referencia al espacio que este contiene. Calcular el volumen de un cuerpo es medir cuánto ocupa mientras que calcular su capacidad es medir cuánto cabe en él.

En la práctica, ambos conceptos son usados indistintamente, ya que tienen unidades equivalentes.

El volumen tiene unidades de longitud al cubo, como por ejemplo:

$m^3,\ cm^3\ ,dm^3,\ pie^3\ , pulg^3$

y la capacidad se suele expresar en litros o unidades derivadas: litro, mililitro, centilitro, etc.

Como mencionamos, hay equivalencia engre los dos grupos de unidades. Entre las más conocidas están:

$1\ lt=1\ dm^3,\ 1\ m^3=1000\ lt$

5 0

3 years ago