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3 years ago

What is the difference between condensation and precipitation

2 answers:

5 0

Answer: The difference between condensation and precipitation is the is the force of gravity that acts on precipitation and draws down the heavy liquid water down to the earth in form of rain.

Explanation: Condensation is the physical change of gaseous phase of water that is water vapour into liquid phase as liquid water an example is fog. Condensation basically depends on temperature and pressure.

Precipitation is the condensation of gaseous phase of water i.e. atmospheric water vapour which falls under gravity from clouds back to the earth in form of drizzle and/or rain. Precipitation depends on the temperature and the concentration of the solution.

Alexxandr [17]3 years ago

3 0

Explanation:

The water cycle is based on three parts;

1. Evaporation

2. Condensation

3. Participation

Condensation:

It is the process in which water vapor changes into liquid water or in other words, it is the transition from the gaseous state to liquid state.

Precipitation:

It is the process in which any liquid or frozen water such as snow that forms in the atmosphere and falls back to the Earth

Condensation depends on temperature and pressure whereas precipitation depends on the temperature and the concentration of the solution.

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An experiment is carried out to measure the extension of a rubber band for different loads.

PtichkaEL [24]

Complete question is;

An experiment is carried out to measure the extension of a rubber band for different loads.

The results are shown in the image attached.

What figure is missing from the table?

Answer:

17.3 cm

Explanation:

The image attached showed values for load, extension and initial length.

Now, the first length there is 15.2 cm and as such it's corresponding extension is 0 because it has no preceding measured length.

The second measured length is 16.2 cm. Since it's initial measured length is 15.2 cm, then the extension has a formula; final length - initial length.

This gives: 16.2 - 15.2 = 1 cm

This corresponds to what is given in the table.

For the next measured length, it is blank but we are given the extension to be 2.1 cm. Now, since the initial measured length is 15.2 cm.

Thus;

2.1 cm = Final length - 15.2 cm

Final length = 15.2 + 2.1

Final length = 17.3 cm

3 0

2 years ago

What graph shape is this what does the snap tell you

vekshin1

The picture is hard to see but if you still need help message me

7 0

2 years ago

Rubric for Grading O Activity3. Answer me! Directions: Compare and contrast the scientific method in philosophy and in science.

emmasim [6.3K]

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2 years ago

Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant

irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

$a_g = \frac{GM}{R^2}$

Here

$M = \text{Mass inside the Orbit of the star}$

$R = \text{Orbital radius}$

$G = \text{Universal Gravitational Constant}$

Mass inside the orbit in terms of Volume and Density is

$M =V \rho$

Where,

V = Volume

$\rho =$ Density

Now considering the volume of the star as a Sphere we have

$V = \frac{4}{3} \pi R^3$

Replacing at the previous equation we have,

$M = (\frac{4}{3}\pi R^3)\rho$

Now replacing the mass at the gravitational acceleration formula we have that

$a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho$

$a_g = \frac{4}{3} G\pi R\rho$

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration. So centripetal acceleration of the star is

$a_c = \frac{4}{3} G\pi R\rho$

At the same time the general expression for the centripetal acceleration is

$a_c = \frac{\Theta^2}{R}$

Where $\Theta$ is the orbital velocity

Using this expression in the left hand side of the equation we have that

$\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2$

$\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}$

$\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R$

Considering the constant values we have that

$\Theta = \text{Constant} \times R$

$\Theta \propto R$

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

6 0

3 years ago

Space-faring astronauts cannot use standard weight scales (since they are constantly in free fall) so instead they determine the

valentinak56 [21]

Answer:

ma = 48.48kg

Explanation:

To find the mass of the astronaut, you first calculate the mass of the chair by using the information about the period of oscillation of the empty chair and the spring constant. You use the following formula:

$T=2\pi\sqrt{\frac{m_c}{k}}$ (1)

mc: mass of the chair

k: spring constant = 600N/m

T: period of oscillation of the chair = 0.9s

You solve the equation (1) for mc, and then you replace the values of the other parameters:

$m_c=\frac{T^2k}{4\pi^2}=\frac{(0.9s)^2(600N/m)}{4\pi^2}=12.31kg$ (2)

Next, you calculate the mass of the chair and astronaut by using the information about the period of the chair when the astronaut is sitting on the chair:

T': period of chair when the astronaut is sitting = 2.0s

M: mass of the astronaut plus mass of the chair = ?

$T'=2\pi\sqrt{\frac{M}{k}}\\\\M=\frac{T'^2k}{4\pi^2}=\frac{(2.0s)^2(600N/m)}{4\pi^2}\\\\M=60.79kg$ (3)

Finally, the mass of the astronaut is the difference between M and mc (results from (2) and (3)) :

$m_a=M-m_c=60.79kg-12.31kg=48.48kg$

The mass of the astronaut is 48.48 kg

3 0

3 years ago