All categories

3 years ago

The atmosphere protects us from _ and _.

Physics

2 answers:

sveticcg [70]3 years ago

7 0

Answer: B. meteorites, UV rays

The earth's atmosphere plays an important role in shielding the space rocks and meteorites enters inside the earth due to collision with the celestial body. Due to high heat and pressure in the mesospheric layer of the atmosphere these meteorites burned up before reaching the biosphere or hydrosphere system of the earth. The UV rays are protected from entering into the earth atmosphere by the ozone layer present in the stratosphere of the atmosphere. Both meteorites and UV rays can negatively effect the human population. Therefore, the atmosphere is an agent which provides protection against them.

Mekhanik [1.2K]3 years ago

5 0

Answer:

B. meteorites, UV rays

Explanation:

Atmosphere of Earth acts like a protective layer around us. It also protects us from:

1. Meteoroid: These are small space rocks revolving around the Sun but cross the orbit of Earth. When the Earth crosses through such areas these rocks get attracted towards Earth due to gravity and move at a very high speed. The atmosphere prevents all the meteoroids from hitting the surface. This happens because the meteoroids get burned in the atmosphere itself due to friction. These burnt meteoroid appear like a shooting star and are called as Meteors. If some part of the rock still survives and land on Earth, it is called as meteorite. If the atmosphere was not there all small/large meteoroid would have hit the Earth.

2. UV rays: UV or ultraviolet rays are a a part of the whole EM spectrum and Sun is a major source of UV rays for us. But UV rays are harmful for us. These rays are blocked by the ozone layer of atmosphere.

You might be interested in

If you wanted to know the location of a vehicle that ran out of gas after taking a zigzag route through the city, which quantity

Papessa [141]

Explanation:

the vehicles displacement, since displacement deals with position

8 0

3 years ago

There are two parallel conductive plates separated by a distance d and zero potential. Calculate the potential and electric fiel

taurus [48]

Answer:

The total electric potential at mid way due to 'q' is $\frac{q}{4\pi\epsilon_{o}d}$

The net Electric field at midway due to 'q' is 0.

Solution:

According to the question, the separation between two parallel plates, plate A and plate B (say) = d

The electric potential at a distance d due to 'Q' is:

$V = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d}$

Now, for the Electric potential for the two plates A and B at midway between the plates due to 'q':

For plate A,

$V_{A} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Similar is the case with plate B:

$V_{B} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Since the electric potential is a scalar quantity, the net or total potential is given as the sum of the potential for the two plates:

$V_{total} = V_{A} + V_{B} = \frac{1}{4\pi\epsilon_{o}}.q(\frac{1}{\frac{d}{2}} + \frac{1}{\frac{d}{2}}$

$V_{total} = \frac{q}{4\pi\epsilon_{o}d}$

Now,

The Electric field due to charge Q at a distance is given by:

$\vec{E} = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d^{2}}$

Now, if the charge q is mid way between the field, then distance is $\frac{d}{2}$ .

Electric Field at plate A, $\vec{E_{A}}$ at midway due to charge q:

$\vec{E_{A}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Similarly, for plate B:

$\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Both the fields for plate A and B are due to charge 'q' and as such will be equal in magnitude with direction of fields opposite to each other and hence cancels out making net Electric field zero.

3 0

3 years ago

A projector is placed on the ground 22 ft. away from a projector screen. A 5.2 ft. tall person is walking toward the screen at a

Stella [2.4K]

Answer:

y = 67.6 feet, y = 114.4/ (22 - 3t)

Explanation:

For this exercise let's use that light travels in a straight line and some trigonometric relationships, the symbols are in the attached diagram

Large triangle Projector up to the screen

tan θ = y / L

For the small triangle. Projector up to the person

tan θ = y₀ / (L-d)

The angle is the same, so we equate the two equations

y₀ / (L -d) = y / L

y = y₀ L / (L-d)

The distance from the screen (d), we look for it with kinematics

v = d / t

d = v t

we replace

y = y₀ L / (L - v t)

y = 5.2 22 / (22 - 3 t)

y = 114.4 (22 - 3t)⁻¹

This is the equation of the shadow height change as a function of time

For the suggested distance the shadow has a height of

y = 114.4 / (22-13)

y = 67.6 feet

7 0

3 years ago

Which of these is emitted during beta decay ?

satela [25.4K]

Answer:

C. a small charged particle.

Explanation:

typically beta radiation emits an electron which is a small negativity charged particle.

hope it helps. :)

4 0

3 years ago

Read 2 more answers

Sari made pancake batter. She has 1 cup of batter left. How much batter did she use?

Naily [24]

She used 1 cup less than the amount that she made when she decided to make pancakes. We have no idea how much she had to begin with.

5 0

3 years ago

The atmosphere protects us from _____ and _____.

The atmosphere protects us from _ and _.