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shutvik [7]
3 years ago
15

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.

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To water the yard you use a hose with a diameter of 3.2 cm. Water flows from the hose with a speed of 1.1 m/s. If you partially
cupoosta [38]

Answer:

The speed of water flow inside the pipe at point - 2 = 34.67 m / sec

Explanation:

Given data

Diameter at point - 1 = 3.2 cm

Velocity at point - 1 = 1.1 m / sec = 110 cm / sec

Diameter at point - 2 = 0.57 cm

Velocity at point - 2 = ??

We know that from the continuity equation the rate of flow is constant inside  a pipe between two points.

Thus

⇒ A_{1} × V_{1} = A_{2} × V_{2}

⇒  \frac{\pi }{4} × d_{1} ^{2} × V_{1} =

⇒  d_{1} ^{2} × V_{1} =  d_{2} ^{2}  × V_{2}

⇒  (3.2)^{2} × 110 = (0.57)^{2} × V_{2}

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⇒ V_{2} = 34.67 m / sec  

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3 0
3 years ago
Two particles with charges +6e and -6e are initially very far apart (effectively an infinite distance apart). They are then fixe
JulijaS [17]

Answer:

\rm EPE_{final}-EPE_{initial}=-1.478\times 10^{-15}\ J.

Explanation:

Given charges are:

\rm q_1 = +6e.\\q_2 = -6e.

The electric potential energy of a charge due to the electric field of another charge is given by

\rm EPE=\dfrac{kq_1q_2}{r}.

where,

  • k = Coulomb's constant, having value = \rm 9\times 10^9\ Nm^2/C^2.
  • r = distance between the charges.

When the charges are infinite distance apart, \rm r = \infty,

\rm EPE_{initial} = \dfrac{kq_1q_2}{r}=0\ J.

When the charges are \rm 5.61\times 10^{-12}\ m apart, \rm r=5.61\times 10^{-12}\ m,

\rm EPE_{final}=\dfrac{kq_1q_2}{r}\\=\dfrac{(9\times 10^9)\times (+6e)\times (-6e)}{5.61\times 10^{-12}}\\=-5.775\ e^2\times 10^{22}.

Here, e is the charge on one electron, such that, \rm e = -1.6\times 10^{-19}\ C.

Therefore,

\rm EPE_{final}=-5.775\times (-1.6\times 10^{-19})^2\times 10^{22} = -1.478\times 10^{-15}\ J.

Thus,

\rm EPE_{final}-EPE_{initial}=-1.478\times 10^{-15}-0=-1.478\times 10^{-15}\ J.

4 0
3 years ago
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