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mote1985 [20]
3 years ago
15

Dose aloe Vera grow in the mesa?

Physics
1 answer:
Zina [86]3 years ago
3 0
there are diffrent species of aloe vera so there is
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Suppose a star has the same apparent brightness as Alpha Centauri A ( 2.7×10−8watt/m2 ) but is located at a distance of 300 ligh
iragen [17]

For  a star that has the same apparent brightness as Alpha Centauri A ( 2.7×10−8watt/m2 is mathematically given as

L=2.7*10^30w

<h3> What is its luminosity?</h3>

Generally, the equation for the luminosity  is mathematically given as

L=4*\pi^2*b

Therefore

L=4*\pi^2*b

L=4* \pi *(2.83*10^{18})*2.7*10^{-8}

L=2.7*10^30

In conclusion, the luminosity

L=2.7*10^30w

Read more about Light

brainly.com/question/25770676

6 0
2 years ago
What is the range of a ball thrown horizontally at 12 m/s if its time of flight is 3.0 s?
vladimir1956 [14]

Answer:

6

Explanation:

because I did this assignment, :) your welcome

Next time do it by yourself, but here's the answer kid

7 0
3 years ago
Read 2 more answers
write aparagraph explaining the difference bln the heat energy in a substances temperature in physics​
iVinArrow [24]

The core difference is that heat deals with thermal energy, whereas temperature is more concerned with molecular kinetic energy. Heat is the transfer of thermal energy, whereas temperature is a property the object exhibits.

5 0
3 years ago
which statement is true about the atomic mass? a. it is never greater than the atomic number b. it always equal the atomic numbe
Shtirlitz [24]
D. it can be greater than the atomic number
5 0
3 years ago
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During a solar eclipse, the Moon is positioned directly between Earth and the Sun. Find the magnitude of the net gravitational f
dezoksy [38]

Answer:

F= 2.3733 x10^{20} N

Explanation:

Let's define the variables to proceed with the operations,

So,

The masses

M_ {sun} = 1.99 * 10^{30} Kg

M_ {Earth} = 5.98 * 10 ^ {24} Kg

M_ {Moon} = 7.36 * 10 ^{22} Kg

Average distances

\bar {x} _ {Sun \rightarrow Earth} = 1.5 * 10 ^ {11} m

\bar {x} _ {Earth \rightarrow Moon} = 3.84 * 10 ^ 8m

Gravitational constant

G = 6.67 * 10^ {-11} \frac {Nm ^ 2} {kg ^ 2}

The formula of the Gravitational Force between the Moon and the Earth would be,

F = \frac {GM_ {Earth} M_ {Moon}} {(\bar {x} _ {Earth \rightarrow Moon}) ^ 2}

F= \frac{(6.67*10^{-11} \frac{Nm^2}{kg^2})(5.98*10^{24}Kg)(7.36*10^{22}Kg)}{(3.84*10^8m)^2}

F = 1.9908 * 10 ^{20} N

This force is in the direction of the earth.

We perform the same process but now between the Sun and the Moon, like this,

F_2 = \frac {GM_ {Sun} M_ {Moon}} {(\bar {x} _ {Sun \rightarrow Earth} - \bar {x} _ {Earth \rightarrow Moon}) ^ 2}

F_2 = \frac{(6.67*10^{-11} \frac{Nm^2}{kg^2})(1.99*10^{30}Kg)(7.36*10^{22}Kg)}{(1.5*10^{11}m-3.84*10^8m)^2}

F_2 = 4.3641*10^-{ 20} N

This force is in the direction of the Sun

The net force must be

F_ {net} = F_2-F

F_ {net} = 2.3733*10^{20} N

This in the direction of the Sun.

8 0
3 years ago
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