QUESTION 3

The atmosphere protects us from _____ and _____.

sveticcg [70]

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.

7 0

3 years ago

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Which of the below is an example of mimicry that enables prey species avoid predation?

Marta_Voda [28]

b..a harmless organism imitating the look of a harmful organism

Explanation:

A harmless organism imitating the look of a harmful organism is one example of mimicry that enables prey species avoid predation.

Prey are smaller and less harmful organisms often hunted by larger organisms usually carnivores.

Mimicry is a form of evolutionary adaptation process in which two organisms of the same specie or different species tends to look alike.

It is a subtle defense mechanism developed by organism over an extended period of time.

learn more:

Adaptation brainly.com/question/11105547

#learnwithBrainly

6 0

3 years ago

A soccer player carries the ball for a distance of 40.0 m in the direction 42.0° west of south. find the westward component of t

Maksim231197 [3]

Actually what the problem meant about the westward component of the ball’s displacement is the horizontal component of the displacement. To help us better understand the problem, I attached a figure of the situation.

We can see from the figure that to solve for the value of the horizontal component, we have to make use of the sin function. That is:

sin θ = side opposite to the angle / hypotenuse of the triangle

sin 42 = x / 40 m

x = (40 m) sin 42

x = 26.77 m

Therefore the ball has a westward displacement of about 26.77 m

3 0

3 years ago

The initial kinetic energy imparted to a 0.020 kg bullet is 1200 J. (a) Assuming it

Lilit [14]

Answer:

(a) Power= 207.97 kW

(b) Range= 5768.6 meter

Explanation:

Given,

Mass of bullet, $m=0.02 kg$

Kinetic energy imparted, $K=1200 J$

Length of rifle barrel, $d=1 m$

(a)

Let the speed of bullet when it leaves the barrel is $v$ .

Kinetic energy, $K=\frac{1}{2} mv^{2}$

$v=\sqrt{\frac{2K}{m} }$

$=\sqrt{\frac{2\times1200}{0.02} }$

$=346.4m/s$

Initial speed of bullet, $u=0$

The average speed in the barrel, $v_a_v_g=\frac{u+v}{2}$

$=\frac{0+346.4}{2} \\=173.2 m/s$

Time taken by bullet to cross the barrel, $t=\frac{d}{v}$

$=\frac{1}{173.2}\\ =0.00577 second$

Power, $P_a_v_g=\frac{W}{t}$

$=\frac{1200}{0.00577} \\=207.97kW$

(b)

In projectile motion,

Maximum height, $H_m=\frac{v^2\sin^2\theta}{2g} \\$

Range, $R=\frac{v^2\sin2\theta}{g}$

given that, $H_m=R$

then, $\frac{v^2\sin^2\theta}{2g}=\frac{v^2\sin2\theta}{g}\\\sin^2\theta=2\sin\theta\cos\theta\\\\\tan\theta=4\\\theta=\tan^-^14\\\theta=75.96^0\\R=\frac{v^2\sin2\theta}{g}\\=\frac{346.4^2\times\sin(2\times75.96)}{9.8}\\5768.6 meter$