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4 years ago

There are 2 questions :)

2 answers:

8 0

For the first question, evaporation is when water evaporates and rises into the air, causing condensation, which turns warm and moist air into a cloud.
For the second question, if it were to precipitate in an area that is low in elevation, there would be no snow, sleet, or hail. There would only be raining, although not a lot. An example of these places is the desert (like Death Valley). If it were to precipitate in an area that is high in elevation, then it would be a different matter altogether. It would rain frequently, and it would also snow. There may also be sleet or hail. An example of these places would be the mountains (like Big Bear Mountain or something). Finally, if it is a level area and is near sea level, than it would neither precipitate frequently or almost never, it would just be like maybe it rained today, and we may not have any more rain for a month, or maybe it might unexpectedly rain for another 5 days in a row. You know what I mean? An example for these places is... let's say Irvine.
I hoped this helped you and have a great day!

zaharov [31]4 years ago

3 0

I can answer the first question: Water rises because of the sun's heat and turns into water vapor. After that, it gathers up and forms a cloud. When the cloud is filled enough with vapor, precipitation occurs.

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*Physical Science* *E2020* *Unit Test*

vladimir2022 [97]

i think its b

hope this helps

5 0

3 years ago

Read 2 more answers

A NASA explorer spacecraft with a mass of 1,000 kg takes off in a positive direction from a stationary asteroid. If the velocity

Georgia [21]

Answer: 10000 kg

Explanation:

The momentum $p$ is given by the following equation:

$p=m.V$ (1)

Where:

$m$ is the mass of the object

$V$ is the velocity of the object

Now, in this case and according the conservation of momentum:

$m_{1}v_{1}+m_{2}v_{2}=m_{1}u_{1}+m_{2}u_{2}$ (2)

Where:

$m_{1}=1000kg$ is the mass of the spacecraft

$m_{2}$ is the mass of the asteroid

$v_{1}=0$ is the initial velocity of the spacecraft

$v_{2}=0$ is the initial velocity of the asteroid (because we are told the asteroid is stationary, as the spacracft is on the sateroid it remains stationary as well)

$u_{1}=250m/s$ is the final velocity of the spacecraft

$u_{2}=-25m/s$ is the final velocity of the asteroid

Rewritting (2):

$0=m_{1}u_{1}+m_{2}u_{2}$ (3)

$0=(1000kg)(250m/s)+m_{2}(-25m/s)$ (4)

Finding $m_{2}$ :

$m_{2}=10000kg$ This is the mass of the asteroid

3 0

3 years ago

Read 2 more answers

An object with a density of 0.85 g/cc is dropped into each of the two beakers shown below. Beaker 1 has a density of 0.5 g/cc. B

Oliga [24]

Answer:

The object will sink in the liquid in beaker 1.

The object will float in the liquid in beaker 2

Explanation:

The density of an object relative to the density of a fluid determines if the object floats or sink in a fluid. The density of a material is the measure of the amount of mass of that material packed into a unit volume of that material.

For the beaker 1, the liquid in this beaker has a density of 0.5 g/cc, which is lesser than the density of the object (0.85 g/cc). This means that the object will add more mass than there should be to the volume of the space it displaces within the field. This results in the object sinking in the fluid.

For beaker 2, the liquid in this beaker has a density of 1 g/cc, which is more than the density of the object (0.85 g/cc). This means that the object will add less mass than there should be to the volume of the space it displaces within the field. This results in the object floating in the fluid.

8 0

4 years ago

A car travels on a straight, level road. (a) Starting from rest, the car is going 38 ft/s (26 mi/h) at the end of 4.0 s. What is

lbvjy [14]

Answer:

$a)9.5\frac{ft}{s^2}\\ b) 12.66\frac{ft}{s^2}$

Explanation:

A body has acceleration when there is a change in the velocity vector, either in magnitude or direction. In this case we only have a change in magnitude. The average acceleration represents the speed variation that takes place in a given time interval.

$a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{38\frac{ft}{s}-0}{4 s- 0}=9.5\frac{ft}{s^2}\\$

$a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{76\frac{ft}{s}-38\frac{ft}{s}}{7 s- 4s}\\a_{avg}=\frac{38\frac{ft}{s}}{3s}=12.66\frac{ft}{s^2}$