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Bogdan [553]
3 years ago
8

Identify impacts of air pollution

Physics
2 answers:
ss7ja [257]3 years ago
3 0

Answer:

Explanation:

High levels of air pollution can cause an increased risk of heart attack, wheezing, coughing, and breathing problems, and irritation of the eyes, nose, and throat. Air pollution can also cause worsening of existing heart problems, asthma, and other lung complications.

Ambient air pollution accounts for an estimated 4.2 million deaths per year due to stroke, heart disease, lung cancer and chronic respiratory diseases. Around 91% of the world's population lives in places where air quality levels exceed WHO limits.

Serious Effects of Pollution on Our Humans and Environment

Environment Degradation. The environment is the first casualty for the increase in pollution weather in air or water. ...

Human Health. ...

Global Warming. ...

Ozone Layer Depletion. ...

Infertile Land.

Please allow me to know if I helped you by clicking thank you, rating or brainly.

Thank you,

Miss Hawaii

svetoff [14.1K]3 years ago
3 0
Long-term health effects from air pollution include heart disease, lung cancer, and respiratory diseases such as emphysema. Air pollution can also cause long-term damage to people's nerves, brain, kidneys, liver, and other organs. Some scientists suspect air pollutants cause birth defects. Air pollution can damage crops and trees in a variety of ways. Ground-level ozone can lead to reductions in agricultural crop and commercial forest yields, reduced growth and survivability of tree seedlings, and increased plant susceptibility to disease, pests and other environmental stresses. Exposure to high levels of air pollution can cause a variety of adverse health outcomes. It increases the risk of respiratory infections, heart disease and lung cancer. Both short and long term exposure to air pollutants have been associated with health impacts. More severe impacts affect people who are already ill.


Hope it can help you:)
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How much heat is needed to raise the temperature of 50.0 g of water by 25.0°C
love history [14]

Answer:

Explanation:

In order to be able to solve this problem, you will need to know the value of water's specific heat, which is listed as

c

=

4.18

J

g

∘

C

Now, let's assume that you don't know the equation that allows you to plug in your values and find how much heat would be needed to heat that much water by that many degrees Celsius.

Take a look at the specific heat of water. As you know, a substance's specific heat tells you how much heat is needed in order to increase the temperature of

1 g

of that substance by

1

∘

C

.

In water's case, you need to provide

4.18 J

of heat per gram of water to increase its temperature by

1

∘

C

.

What if you wanted to increase the temperature of

1 g

of water by

2

∘

C

? You'd need to provide it with

increase by 1

∘

C



4.18 J

+

increase by 1

∘

C



4.18 J

=

increase by 2

∘

C



2

×

4.18 J

To increase the temperature of

1 g

of water by

n

∘

C

, you'd need to supply it with

increase by 1

∘

C



4.18 J

+

increase by 1

∘

C



4.18 J

+

...

=

increase by n

∘

C



n

×

4.18 J

Now let's say that you wanted to cause a

1

∘

C

increase in a

2-g

sample of water. You'd need to provide it with

for 1 g of water



4.18 J

+

for 1 g of water



4.18 J

=

for 2 g of water



2

×

4.18 J

To cause a

1

∘

C

increase in the temperature of

m

grams of water, you'd need to supply it with

for 1 g of water



4.18 J

+

for 1 g of water



4.18 J

+

,,,

=

for m g of water



m

×

4.18 J

This means that in order to increase the temperature of

m

grams of water by

n

∘

C

, you need to provide it with

heat

=

m

×

n

×

specific heat

This will account for increasing the temperature of the first gram of the sample by

n

∘

C

, of the the second gram by

n

∘

C

, of the third gram by

n

∘

C

, and so on until you reach

m

grams of water.

And there you have it. The equation that describes all this will thus be

q

=

m

⋅

c

⋅

Δ

T

, where

q

- heat absorbed

m

- the mass of the sample

c

- the specific heat of the substance

Δ

T

- the change in temperature, defined as final temperature minus initial temperature

In your case, you will have

q

=

100.0

g

⋅

4.18

J

g

∘

C

⋅

(

50.0

−

25.0

)

∘

C

q

=

10,450 J

Rounded to three sig figs and expressed in kilojoules, t

Explanation:

3 0
3 years ago
Read 2 more answers
a lamp or bulb is marked 12 volt 240 Watts how many joules does it consume in one hour and what is the current that passes throu
valkas [14]

Answer:

1 watt is 1 Joule per second. 240 watt would then be 240 joules per second. How many seconds are there in an hour? Current formula is P= IV , P is 240(not the power in an hour because the current is calculated in seconds), and V is 12V. Then you will get I, the current.

5 0
3 years ago
If you run at 8m/s for 15 minutes how far will you go
sashaice [31]

Answer: 7200 m

Explanation: The solution is, first convert 15 minutes to seconds.

15 mins x 60 s / 1 min = 900 s

Use the formula for speed which is v= d/t then derive for d.

d = vt

= 8 m/s ( 900s)

= 7200 m

6 0
3 years ago
Read 2 more answers
What can happen to solar radiation when it enters atmosphere​
grandymaker [24]

Answer:

The solar radiation is first intercepted by Earth's atmosphere, just a small part of the radiation is absorbed by gases such as water vapor. Some of the radiation is reflected back to space by the clouds and Earth's surface.

3 0
3 years ago
Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equ
Irina-Kira [14]

Answer:

Magnitude of the net force on q₁-

Fn₁=1403 N

Magnitude of the net force on q₂+

Fn₂= 810 N

Magnitude of the net force on q₃+

Fn₃= 810 N

Explanation:

Look at the attached graphic:

The charges of the same sign exert forces of repulsion and the charges of   opposite sign exert forces of attraction.

Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:

F= (k*q*q)/(d)²

F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N

Magnitude of the net force on q₁-

Fn₁x= 0

Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N

Fn₁=1403 N

Magnitude of the net force on q₃+

Fn₃x= 810- 810 cos 60° = 405 N

Fn₃y= 810*sin 60° = 701.5 N

Fn_{3} = \sqrt{405^{2}+701.5^{2}  }

Fn₃ = 810 N

Magnitude of the net force on q₂+

Fn₂ = Fn₃ = 810 N

6 0
3 years ago
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