Ask question

Ask question

All categories

2 years ago

10

Smog is a type of air pollution. It is a mixture of fog and smoke or other airborne pollutants such as exhaust fumes. What is a

result of smog that effects the quality of life for organisms? A. acid rain B. floods C. all of these

2 answers:

Tasya [4]2 years ago

7 0

The answer is A. Acid rain

Akimi4 [234]2 years ago

5 0

the answer acid rain

You might be interested in

The Big Bang Theory was supported by the discovery of the

Alinara [238K]

Two major scientific discoveries provide strong support for the Big Bang theory: • Hubble's discovery in the 1920s of a relationship between a galaxy's distance from Earth and its speed; and • the discovery in the 1960s of cosmic microwave background radiation. Please give brainliest

7 0

2 years ago

Read 2 more answers

Suppose you are planning a trip in which a spacecraft is to travel at a constant velocity for exactly six months, as measured by

bixtya [17]

Answer:

v = 0.99 c = 2.99 x 10⁸ m/s

Explanation:

From the special theory of relativity:

$t = \frac{t_o}{\sqrt{1-\frac{v^2}{c^2} } }\\$

where,

v = speed of travel = ?

c = speed of light = 3 x 10⁸ m/s

t = time measured on earth = 90 years

t₀ = time measured in moving frame = 6 months = 0.5 year

Therefore,

$90\ yr = \frac{0.5\ yr}{\sqrt{1-\frac{v^2}{c^2} } } \\\\\\\sqrt{1-\frac{v^2}{c^2}} = \frac{0.5\ yr}{90\ yr}\\ 1-\frac{v^2}{c^2} = 0.00003086\\\frac{v^2}{c^2} = 1-0.00003086\\$

<u>v = 0.99 c = 2.99 x 10⁸ m/s</u>

8 0

2 years ago

In a vacuum, two particles have charges of q1 and q2, where q1 = +3.11 μC. They are separated by a distance of 0.241 m, and part

lesya [120]

Answer:

Charge of particle 2, $q_2=-7.13\ \mu C$

Explanation:

Given that,

Charge 1, $q_1=3.11\ \mu C=3.11\times 10^{-6}\ C$

The distance between charges, r = 0.241 m

Force experienced by particle 1, F = 3.44 N

We need to find the magnitude of electric charge 2. It can be calculated using formula of electrostatic force. It is given by :

$F=k\dfrac{q_1q_2}{r^2}$

$q_2=\dfrac{Fr^2}{kq_1}$

$q_2=\dfrac{3.44\times (0.241)^2}{9\times 10^9\times 3.11\times 10^{-6}}$

$q_2=7.13\times 10^{-6}\ C$

or

$q_2=7.13\ \mu C$

So, the magnitude of electric charge 2 is $q_2=7.13\ \mu C$ . Since, the force is attractive then the magnitude of charge 2 must be negative.

5 0

2 years ago

Which of the following will not reduce water pollution?

Rina8888 [55]

Answer:

A. Using artificial fertilizers instead of composting

Explanation:

Composting is actually more better than fertilizers usually, as it takes natural waste and with help of nature, returning the nutrients into soil once again. However, Artificial fertilizers usually uses chemicals to ensure that weed does not grow, and may contain other "unnatural" ingredients to ensure that only the crop grows on the field. Because of the usage of non-natural ingredients, any run-off can prove to be detrimental to any animals that eat any of the run-off infected material, or drink the polluted water.

~

7 0

2 years ago

Read 2 more answers

If you place 48 cm object 15 m in front of a convex mirror of focal length 5 m, what is the magnification of the image?

Arisa [49]

Answer:

m = 0.25

Explanation:

Given that,

Object distance, u = -15cm

Height of the object, h = 48

Focal length, f = cm

We need to find the magnification of the image.

Let v is the image distance. Using mirror's equation.

$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}\\\\=\dfrac{1}{5}-\dfrac{1}{(-15)}\\\\=3.75\ cm$

Magnification,

$m=\dfrac{-v}{u}\\\\=\dfrac{-3.75}{-15}\\\\=0.25$

Hence, the magnification of the image is 0.25.

7 0

2 years ago