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3 years ago

cumulonimbus clouds often form along cold forms. which type of weather is most likely to occur along a cold front?

1 answer:

7 0

A cold front phenomenon characterized by cold air pushing into much warmer air mass. The cold air being more dense than warm air is able to effectively push it forward usually at a fast pace.
When the cold front is passing through an area, there are gusts of wind which usually sweep right across the ground and raise the dust in dry areas, there is a sudden drop in temperature with moderate to heavy rain sometimes with thunder, lightning and hail.
The warm air moving ahead of the cold front produces cumulus or cumulonimbus clouds with thunderstorms as atmospheric pressure falls then slowly rises again after the cold front has passed.
Temperatures are noticeably cooler in a given area after a cold front has passed, which is followed by replacement of cumulus clouds by stratus and stratocumulus clouds and clear skies.

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Artificial gravity is a must for any space station if humans are to live there for any extended length of time. Without artifici

wariber [46]

Answer:

Hi myself Shrushtee.

Explanation:

Artificial gravity is a must for any space station if humans are to live there for any extended length of time. Without artificial gravity, human growth is stunted and biological functions break down. An effective way to create artificial gravity is through the use of a rotating enclosed cylinder, as shown in the figure. Humans walk on the inside edge of the cylinder, which is sufficiently large (diameter of 2235 meters) that its curvature is not readably noticeable to the inhabitants. (The space station in the figure is not drawn to the scale of the human.) Once the space station is rotating at the necessary speed, how many minutes would it take the space station to make one revolution?

The distance traveled by the man in one revolution is simply the circumference of the space station, C = 2p R. From this result, you should be able to deduce the time it takes for the space station to sweep out a complete revolution.

<h2>Please mark me as brainleist</h2>

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3 years ago

Please help...........................

Marianna [84]

Hi!

Neutrons are neutral, which means they don't exactly have an electrical charge. It's because of this neutral charge that it is represented with a '0'.

On the other hand, protons and electrons do have electrical charges. Electrons flow around the outside of the nucleus, with a negative charge.

Protons are stored in the nucleus with the neutrons, holding a positive charge.

Hopefully, this helps! =)

6 0

4 years ago

Read 2 more answers

A bird, accelerating from rest at a constant rate, experiences a displacement of 28 m in 11 s. what is the final velocity after

frutty [35]

Let's see what variables we've got first. Hmmm. We have:

Displacement, d = 28 m
Time taken, t = 11 s
Initial velocity, u = 0 m/s (at rest)

And now we need to find the final velocity, v. Among the 4 (or 5) equations of motions, there's no equation that will let us simply plug in the values and give an answer sigh. But fear not! We'll do it in steps.

I'm going to pick one of the motion equation to find more information:

$d = ut + \frac{1}{2} a {t}^{2}$

I know everything except for a in this one, so I I'll use this! After plugging in values, I get a = 0.4628 m/s^2.

Now I'm going to use another motion equation that has v in it because that needs to be solved!

$v = u + at$

Now I know everything except dial velocity v. Nice!

v = 0 + (0.4628)(11)

4 0

3 years ago

What are the density, specific gravity and mass of the air in a room whose dimensions are 4 m * 6 m * 8 m at 100 kPa and 25 C.

Volgvan

Answer:

Density = 1.1839 kg/m³

Mass = 227.3088 kg

Specific Gravity = 0.00118746 kg/m³

Explanation:

Room dimensions are 4 m, 6 m & 8 m. Thus, volume = 4 × 6 × 8 = 192 m³

Now, from tables, density of air at 25°C is 1.1839 kg/m³

Now formula for density is;

ρ = mass(m)/volume(v)

Plugging in the relevant values to give;

1.1839 = m/192

m = 227.3088 kg

Formula for specific gravity of air is;

S.G_air = density of air/density of water

From tables, density of water at 25°C is 997 kg/m³

S.G_air = 1.1839/997 = 0.00118746 kg/m³

4 0

3 years ago

To determine an athlete's body fat, she is weighed first in air and then again while she's completely underwater. it is found th

il63 [147K]

The difference in weight is due to the displacement of water (the buoyancy of water is acting on the athlete thus giving her smaller weight).

The amount of weight displaced or the amount of buoyant force is:

Fb = 690 N - 48 N

Fb = 642 N

From newtons law, F = m*g. Using this formula, we can calculate for the mass of water displaced:

m of water displaced = 642N / 9.8m/s^2

m of water displaced = 65.5 kg

Assuming a water density of 1 kg/L, and using the formula volume = mass/density:

V of water displaced = 65.5kg / 1kg/L = 65.5 L

The volume of water displaced is equal to the volume of athlete. Therefore:

V of athlete = 65.5 L

The mass of athlete can also be calculated using, F = m*g

m of athlete = 690 N/ 9.8m/s^2

m of athlete = 70.41 kg

Knowing the volume and mass of athlete, her average density is therefore:

average density = 70.41 kg / 65.5 L

average density = 1.07 kg/L = 1.07 g/mL

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3 years ago

Read 2 more answers