All categories

3 years ago

When water in clouds undergoes condensation, it can lead to?

1 answer:

4 0

It will lead to rain. I know this because as the clouds move over water sources like oceans, lakes, and rivers, the water evaporates and rises. The water then liquefies into little water droplets. As the cloud moves over more water the droplets get scrunched up and get bigger over time and soon they get heavy and gravity pulls the droplets down to earth as rain. The End.

You might be interested in

Which energy resources are found above the Earth's surface?

OverLord2011 [107]

Answer:

Hey

Your answer would be B) B) solar, wind, biomass

Hydroelectric, uranium Geothermal=bellow earth's surface.

Coal, Oil, Natural gas=bellow earth's surface.

4 0

3 years ago

. Suppose the mass of a fully loaded module in which astronauts take off from the Moon is 1.00 × 104 kg. The thrust of its engin

Viktor [21]

Answer:

a) The module's acceleration in a vertical takeoff from the Moon will be $1.377 \frac{m}{s^2}$

b) Then we can say that a thrust of $3*10^{4} N$ won't be able to lift off the module from the Earth because it's smaller than the module's weight ( $9.8 *10^{4} N$ ).

Explanation:

a) During a vertical takeoff, the sum of the forces in the vertical axis will be equal to mass times the module's acceleration. In this this case, the thrust of the module's engines and the total module's weight are the only vertical forces. (In the Moon, the module's weight will be equal to its mass times the Moon's gravity acceleration)

$T-(m*g)=m*a$

Where:

$T=$ thrust $=3 *10^{4} N$

$m=$ module's mass $=1 *10^{4} N$

$g=$ moon's gravity acceleration $=1.623 \frac{m}{s^2}$

$a=$ module's acceleration during takeoff

Then, we can find the acceleration like this:

$a=\frac{T}{m} -g=\frac{3*{10}^4 N}{1*{10}^4 kg}-1.623\frac{m}{s^2}$

$a=1.377 \frac{m}{s^2}$

The module's acceleration in a vertical takeoff from the Moon will be $1.377 \frac{m}{s^2}$

b) To takeoff, the module's engines must generate a thrust bigger than the module's weight, which will be its mass times the Earth's gravity acceleration.

$weight=m*g=(1*{10}^4 kg)*(9.8 \frac{m}{s^2})=9.8 *10^{4} N$

Then we can say that a thrust of $3*10^{4} N$ won't be able to lift off the module from the Earth because it's smaller than the module's weight ( $9.8 *10^{4} N$ ).

8 0

4 years ago

What equation describes conservation of charge?

Phantasy [73]

Answer:

The equation which describes conservation of charge is $Q_{initial} - Q_{final } = 0$

Explanation:

The law of conservation charge states that for an isolated system that sum of initial charges is equal to sum of final charges, that is the total charge is conserved.

let the sum of initial charges = $Q_{initial}$

let the sum of the final charges = $Q_{final}$

$Q_{initial } = Q_{final}\\\\Q_{initial } - Q_{final} = 0$

Therefore, the equation which describes conservation of charge is $Q_{initial} - Q_{final } = 0$

6 0

3 years ago

Based on the idea of isostasy, which of these statements is NOT correct?

Wewaii [24]

The formation ice sheets cause the Earth's surface to sink so much that when the ice melts, the crust and mantle does not rebound.

Explanation:

The above-mentioned statement is the only false statement

During the Ice ages, large continental parts were covered under thick ice sheets. This imposed a superincumbent additional load on the Earth's crust.

Isostasy refers to the balance that exists between parts of Earth's mantle and crust. Isostatic adjustments are a common phenomenon. As a result of isostatic adjustments, several parts of the world have risen by as much as 900 feet as an adjustment to superincumbent load since ice ages. E.g. Parts of Scandinavian countries have witnessed these activities and also undergoing this process.

7 0

4 years ago

An object 4 cm high is placed 20 cm in front of a convex lens of focal length 12 cm. What is the position and height of the imag

aleksandrvk [35]

It would be 12cm

hope this helps

6 0

3 years ago