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g100num [7]
3 years ago
6

If Earth were completely blanketed with clouds and we couldn’t see the sky, could we learn about the realm beyond the clouds? Wh

at forms of radiation might penetrate the clouds and reach the ground?
Physics
1 answer:
Fudgin [204]3 years ago
4 0

The definition of waves that propagate through electric fields is called electromagnetic waves. The earth, despite being covered with clouds, can be 'affected' because waves such as sunlight or the moon have the ability to penetrate and be visible to the inhabitants of the earth. Microwaves and radio waves would be less affected by the clouds that cover the Earth.

Through these waves, you can know that there is beyond the clouds.

Ultraviolet light, microwaves and radio waves are the radiations that penetrate through the clouds and reach the Earth's surface.

Therefore, the answer is Yes, ultraviolet light, microwaves and radio waves are the forms of radiation that penetrate and reach the ground.

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A basketball player jumps straight up for a ball. To do this, he lowers his body 0.310 m and then accelerates through this dista
Nastasia [14]

Answer:A)u =4.295m/s  , B)a = 29.746m/s²   C) F=3,153N

Explanation:

Using the kinematic expression  

v² = u² - 2as

where

u = initial velocity

v = final velocity

s = distance

g = acceleration due to gravity .

Given that he reaches a height of 0.940 m above the floor,

the final velocity  = 0

Here, acceleration due to gravity is acting in  opposite the initial direction of motion. So, a=-9.81 m/s.

v² = u² + 2as

0² - u² = 2 (- 9.81) × 0.940

- u² = 2 × - 9.81 × 0.920

- u² = -18.4428

cancelling the minus in both sides , we have that  

u² = 18.4428

u = √18.4428

u =4.295m/s

(b) His acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.310 m. m/s2

Using v² = u² + 2as

where u = initial speed of basketball player before lengthening = 0 m/s,

v = final speed of basketball player after lengthening =  4.295m/s,

a = acceleration while  straightening his legs

s = distance moved during lengthening = 0.310m

v² = u² + 2as  

 a = (v² - u²)/2s

a = (4.29m/s)² - (0 m/s)²)/(2 × 0.310m)

a = (18.4428 m²/s² - 0 m²/s²)/(0.62 m)

a = (18.4428 m²/s²/(0.62 m)

a = 29.746m/s²

c) The force (in N) he exerts on the floor to do this, given that his mass is 106 kg. N

Force= mass x acceleration.

F = 106 kg X 29.746m/s²

 F = 3,153.076 rounded to  3,153N

8 0
3 years ago
Are zebra fish schooling fish?
Alex73 [517]

Answer:

they r schooling fish

Explanation:

they need to be kept in groups of 5.

5 0
3 years ago
Read 2 more answers
Through what process is carbon pulled from the atmosphere in the carbon cycle
Andrei [34K]

Carbon is pulled from the atmosphere in the carbon cycle through the process of photosynthesis. Details about photosynthesis can be found below.

<h3>What is photosynthesis?</h3>

Photosynthesis is the process whereby green plants obtain their nutrition by utilizing energy from sunlight.

Green plants absorb carbon in the form of carbon dioxide from the atmosphere and use it in the photosynthetic process.

This means that one way that carbon is removed from the atmosphere during the carbon cycle is through photosynthesis.

Learn more about photosynthesis at: brainly.com/question/1388366

#SPJ1

7 0
3 years ago
Which best explains why sound travels faster through a solid
Masteriza [31]
Its because the molecules in the solid structures are very close to each other and rigidly packed, thus due to this quantum structure they have pretty awesome speed of sound in them
8 0
3 years ago
Read 2 more answers
Simplified. The absorption spectra of ions have been used to identify the presence of the elements in the atmospheres of the sun
Nookie1986 [14]

Answer:

The answer is "3.83 \times 10^9 \ m"

Explanation:

Z=2, so the equation is E= \frac{-4B}{n^2}

Calculate the value for E when:  

n=2 and n=9

The energy is the difference in transformation, name the energy delta E Deduct these two energies  

In this transition, the wavelength of the photon emitted is:

\Delta E=2.18 \times  10^{-18} ( \frac{1}{4}- \frac{1}{81})

\lambda = \frac{h c}{\Delta E}

h ( Planck's\  constant) = 6.62 \times  10^{(-34)} \ Js \\\\ speed \ of \ light = 3 \times 10^{8} \ \frac{m}{s}\\\\= \frac{6.62 \times 10^{(-34)} \times 3 \times 10^ {8}}{2.18 \times  10^{-18}} (\frac{1}{4}- \frac{1}{81}) \\\\=3.83 \times 10^9 \ m\\\\

6 0
3 years ago
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