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Katarina [22]
2 years ago
7

Name at least two clouds types in each level, with their characteristics:

Physics
2 answers:
sergejj [24]2 years ago
6 0

Answer:

High-Level Clouds

Cloud types include: cirrus and cirrostratus.

High-level clouds form above 20,000 feet (6,000 meters) and since the temperatures are so cold at such high elevations, these clouds are primarily composed of ice crystals. High-level clouds are typically thin and white in appearance, but can appear in a magnificent array of colors when the sun is low on the horizon.

Mid-Level Clouds

Cloud types include: altocumulus, altostratus.

The bases of mid-level clouds typically appear between 6,500 to 20,000 feet (2,000 to 6,000 meters). Because of their lower altitudes, they are composed primarily of water droplets, however, they can also be composed of ice crystals when temperatures are cold enough.

Low-level Clouds

Cloud types include: nimbostratus and stratocumulus.

Low clouds are of mostly composed of water droplets since their bases generally lie below 6,500 feet (2,000 meters). However, when temperatures are cold enough, these clouds may also contain ice particles and snow.

Sholpan [36]2 years ago
5 0

Answer:

Low- Level:  Cumulus and stratus

Cumulus have tops that are rounded, puffy, and a brilliant white when sunlit, while their bottoms are flat and relatively dark.

Stratus clouds hang low in the sky as a flat, featureless, uniform layer of grayish cloud.

Mid-Level:  Altocumulus and nimbostratus

Altocumulus have white or gray patches that dot the sky in large, rounded masses or clouds that are aligned in parallel bands.

Nimbostratus clouds cover the sky in a dark gray layer. They are thick enough to blot out the sun

High-Level: Cirrus and cirrocumulus

Cirrus are thin, white, wispy strands of clouds that streak across the sky. They are made up of tiny ice crystals rather than water droplets.

Cirrocumulus clouds are small, white patches of clouds often arranged in rows that live at high altitudes and are made of ice crystals.

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Where do most stars fall on the Hertzsprung-Russell diagram?
Alex_Xolod [135]

Answer:

Most of the stars occupy the region in the diagram along the line called the main sequence. During the stage of their lives in which stars are found on the main sequence line, they are fusing hydrogen in their cores.

7 0
2 years ago
A 34n force is applied to a 213kg mass how much does the mass accelerate
motikmotik
We Know, F = m*a
Here, F = 34 N
m = 213 Kg

Substitute their values in the equation,
34 = 213 * a
a = 34/213
a = 0.159 m/s²

So, your final answer & the acceleration of the object would be 0.159 m/s²

Hope this helps!
3 0
3 years ago
Please help me , I also have to show work on paper
kramer

Answer:

Choose B

Explanation:

Hope Can I help you

7 0
3 years ago
Suppose a certain car supplies a constant deceleration of A meter per second per second. If it is traveling at 90km/hr. When. th
aksik [14]

Answer:

i)-6.25m/s

ii)18 metres

iii)26.5 m/s or 95.4 km/hr

Explanation:

Firstly convert 90km/hr to m/s

90 × 1000/3600 = 25m/s

(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)

0 = (25)^2 + 2A(50)

0 = 625 + 100A....then moved the other value to one

-625 = 100A

Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)

(ii) Firstly convert 54km/hr to m/s

In which this is 54 × 1000/3600 = 15m/s

then apply the same formula as that in (i)

0 = (15)^2 + 2(-6.25)s

-225 = -12.5s

Hence the stopping distance = 18metres

(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question

0 = u^2 + 2(-6.25)(56)

u^2 = 700

Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s

In km/hr....26.5 × 3600/1000 = 95.4 km/hr

3 0
3 years ago
Assuming that Albertine's mass is 60.0 kg , for what value of μk, the coefficient of kinetic friction between the chair and the
makvit [3.9K]

Answer:

\mu_k=0.101

Explanation:

It is given that,

Mass of Albertine, m = 60 kg

It can be assumed, the spring constant of the spring, k = 95 N/m

Compression in the spring, x = 5 m

A glass sits 19.8 m from her outstretched foot, h = 19.8 m

When she just reach the glass without knocking it over, a force of friction will also act on it. Using the conservation of energy for the spring mass system such that,

\dfrac{1}{2}kx^2=\mu_k mgh

\mu_k=\dfrac{kx^2}{2mgh}

\mu_k=\dfrac{95\times (5)^2}{2\times 60\times 9.8\times 19.8}

\mu_k=0.101

So, the coefficient of kinetic friction between the chair and the waxed floor is 0.101. Hence, this is the required solution.

3 0
3 years ago
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