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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.

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Answers can be seen below

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First we must explain the essential when we clear equations, and that is that if the term we need to clear is accompanied by other terms that are being added up, then those terms go to the other side of the equation to subtract if those terms are subtracting, then they go to the other side to add, if those terms are found multiplying then they go to the other side of the equation to divide and if those other terms are found dividing then they go to the other side of the equation to multiply.

(Primero debemos explicar lo esencial cuando despejamos ecuaciones, y es que si el término que necesitamos despejar va acompañado de otros términos que se están sumando, entonces esos términos van al otro lado de la ecuación para restar si esos términos están restando, luego van al otro lado para sumar, si esos términos se encuentran multiplicando luego van al otro lado de la ecuación a dividir, y si esos términos se encuentran dividiendo, pasan al otro lado de la ecuación a multiplicar.)

1 )

$t=\frac{v}{a}$ ; $d=s*(t-t_{0} )$

$k=\frac{2*U}{x^{2} }$ ; $T_{2}=\frac{P_{2}*V_{2}*T_{1} }{P_{1}*V_{1} } \\$

$L=\frac{F}{\pi*r*P}$ ; $d=\frac{w}{F*cos(o)}$

$t^{2}=\frac{2*x}{g} ; V_{2}=\frac{A_{1}*V_{1} }{A_{2} } \\$

$h=\frac{V}{\pi *r^{2} } ; r=\frac{t}{F*sin(o)}$

$h=\frac{m}{(1/2)*\pi *r^{2} } ; h_{2}=\frac{F_{2}*(1/2)*b_{1} *h_{1} }{F_{1}*(1/2)*b_{2}*h_{2} }$

$b=\frac{mg-ma}{v}; m=\frac{F+kx}{g*cos(o)}$

$a=\frac{v-v_{o} }{t} ; u=\frac{m_{1}+m_{2} }{M}$

$v_{o}=\frac{x-\frac{1}{2}*a*t^{2} }{t} ; F=\frac{W+uNd}{d*cos(o)}$

10)

$h=\frac{E-\frac{1}{2}*m*v^{2} }{mg} ; v_{2} ^{2} = \frac{Dk-\frac{1}{2} m*v_{1}^{2} }{\frac{1}{2}m }$

11)

$N=\frac{mg*sin(o)-F}{u} ; x^{2}=\frac{W+\frac{1}{2}k*x_{1}^{2} }{\frac{1}{2}*k }$

12)

$x=x_{o} +\frac{v^{2-v_{o}^{2} } }{2a} ; m=\frac{P*A-F_{1}-F_{2} }{g}$

13)

$x_{o} = x-\frac{F}{k} ; u=\frac{cos(o)-\frac{a}{g} }{sin(o)}$

14)

$t=\frac{d}{v} +t_{o}$ ; $t_{o} = t-(\frac{v-v_{o} }{a} )$

15)

$F_{2}=\frac{W-F_{1} *d}{d}+F_{3} ; v_{2}^{2}=v_{1}^{2}+\frac{2*Dk}{m}$

16)

$y_{1}=y-\frac{u}{mg} ; x^{2} = \frac{2W}{k}+x_{o} ^{2}$

7 0

3 years ago

Steam in a heating system flows through tubes whose outer diameter is 5 cm and whose walls are maintained at a temperature of 13

svet-max [94.6K]

Answer:

5945.27 W per meter of tube length.

Explanation:

Let's assume that:

Steady operations exist;
The heat transfer coefficient (h) is uniform over the entire fin surfaces;
Thermal conductivity (k) is constant;
Heat transfer by radiation is negligible.

First, let's calculate the heat transfer (Q) that occurs when there's no fin in the tubes. The heat will be transferred by convection, so let's use Newton's law of cooling:

Q = A*h*(Tb - T∞)

A is the area of the section of the tube,

A = π*D*L, where D is the diameter (5 cm = 0.05 m), and L is the length. The question wants the heat by length, thus, L= 1m.

A = π*0.05*1 = 0.1571 m²

Q = 0.1571*40*(130 - 25)

Q = 659.73 W

Now, when the fin is added, the heat will be transferred by the fin by convection, and between the fin and the tube by convection, thus:

Qfin = nf*Afin*h*(Tb - T∞)

Afin = 2π*(r2² - r1²) + 2π*r2*t

r2 is the outer radius of the fin (3 cm = 0.03 m), r1 is the radius difference of the fin and the tube ( 0.03 - 0.025 = 0.005 m), and t is the thickness ( 0.001 m).

Afin = 0.006 m²

Qfin = 0.97*0.006*40*(130 - 25)

Qfin = 24.44 W

The heat transferred at the space between the fin and the tube will be:

Qspace = Aspace*h*(Tb - T∞)

Aspace = π*D*S, where D is the tube diameter and S is the space between then,

Aspace = π*0.05*0.003 = 0.0005

Qspace = 0.0005*40*(130 - 25) = 1.98 W

The total heat is the sum of them multiplied by the total number of fins,

Qtotal = 250*(24.44 + 1.98) = 6605 W

So, the increase in heat is 6605 - 659.73 = 5945.27 W per meter of tube length.

5 0

3 years ago

Monitoring cash flow and keeping track of the bottom line of a business are responsibilities for whom?

lina2011 [118]

Answer:
A financial manager.

Explanation:
This is because a financial manager oversees the financial operations of a company. Generally, a financial manager assumes accounting responsibilities for the company. A financial manager is responsible for planning and managing the company's financial resources.

Side note:
Hope this helps!
Please give Brainliest!

6 0

2 years ago

The atmosphere protects us from _____ and _____.

The atmosphere protects us from _ and _.