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Galina-37 [17]
4 years ago
11

Moving force of air flows through areas of high pressure to areas of low pressure

Physics
2 answers:
vazorg [7]4 years ago
7 0

The answer is true i hope this helps

Anvisha [2.4K]4 years ago
7 0

It's true your welcome

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The change in state of a liquid to a gas​
ioda

evaporation (if that's to fill in the blanks question)

6 0
3 years ago
Read 2 more answers
If the moon were twice as far from years as it is now the following would be true
zubka84 [21]
The question is incomplete.

The distance between the Moon and Earth influences: 1) the attractive gravitational force between them, 2) the tides, 3) the eclipses, 4) the period of each full turn of the moon around the Earth.

Assuming the question refers to the gravitational attraction, we must use the fact that, as per, Newton's Universal Gravitaional Law, the attractive force between the two bodies is inversely related to the square distance that separates them.

Then, if the Moon were twice as far, the gravitational pull would be one fourth (1/4) of actual pull.

7 0
4 years ago
When a nerve cell depolarizes, charge is transferred across the cell membrane, changing the potential difference. For a typical
laila [671]

Answer:

I = 18 x 10⁻⁹ A = 18 nA

Explanation:

The current is defined as the flow of charge per unit time. Therefore,

I = q/t

where,

I = Average Current passing through nerve cell

q = Total flow of charges through nerve cell

t = time period of flow of charges

Here, in our case:

I = ?

q = (9 pC)(1 x 10⁻¹² C/1 pC) = 9 x 10⁻¹² C

t = (0.5 ms)(1 x 10⁻³ s/1 ms) = 5 x 10⁻⁴ s

Therefore,

I = (9 x 10⁻¹² C)/(5 x 10⁻⁴ s)

<u>I = 18 x 10⁻⁹ A = 18 nA</u>

6 0
3 years ago
During a snowball fight two balls with masses of 0.4 kg and 0.6 kg, respectively, are thrown in such a manner that they meet hea
valkas [14]
Considering conservation of momentum;
m1v1 + m2v2 = m3v3

In which,
m1 = mass of snowball 1 = 0.4 kg
v1 = velocity of snowball 1 = 15 m/s
m2 = mass of snowball 2 = 0.6 kg
v2 = velocity of snow ball 2 = 15 m/s
m3 = combined mass = 1 kg
v3 = velocity after comination
Therefore;
0.4*15 + 0.6*15 = 1*v3
v3 = 6+9 = 15 m/s
KE = 1/2mv^2

Then,
KE1 = 1/2*0.4*15^2 = 45 J
KE2 = 1/2*0.6*15^2 = 67.5 J
KE3 = 1/2*1*15^2 = 112.5 J

Therefore, KE3 (kinetic energy after collision) = K1+K2 {kinetic energy before collision). And thus it is 100%.
3 0
3 years ago
¿Por qué si cargas a uno de tus compañeros por cierto tiempo no estás realizando un trabajo mecánico?
VladimirAG [237]

Answer:

I will answer this in English, we can translate it to:

Why if you charge a mate by an amount of time you are not doing work?

This happens because work is defined as the displacement done by a force:

W = d*F

where W is work, d is the distance, and F is the force.

This means that the amount of time that you are charging your mate does not affect the mechanical work, the only time that you are doing work is when you are lifting him.

4 0
4 years ago
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