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4 years ago

How do greenhouse gases such as CO2 and N2O contribute to an increase in Earth’s atmospheric temperature?

Physics

2 answers:

EastWind [94]4 years ago

4 0

Answer:

By trapping the excess heat in the Earth's Atmosphere

Explanation:

When Sun light and other radiations fall on Earth a part of it is reflected back. Greenhouse gases such as CO₂ andN₂O traps the reflected heat and maintain the temperature on Earth. This is required to maintain a temperature that is suitable for survival of life. If these gases are not present in the atmosphere, Earth will be too cold to live. At the same time if these gases are present in large amount, they will trap more heat and thus increase the overall temperature of the Earth resulting in Global Warming.

Due to various reasons the amount of such gases has been increasing in our atmosphere in turn increasing Earth's temperature.

konstantin123 [22]4 years ago

3 0

CO2 and N2O keep the energy that gets to Earth from the sun inside the atmosphere. Without greenhouse gases, our planet would be too cold. But due to the recent increase in greenhouse gases, more energy released from the sun is contained in the Earth, heating it up.

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Which device in a car helps to protect riders from the effects of inertia in the event of an acident

Schach [20]

The seatbelt, as it prevents you from flying out of your window.

if this was the appropriate answer make sure to mark as the brainliest!
-procklown

5 0

4 years ago

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An athlete runs 200 m by running 100 m in a straight line and then turning around and running 100m back to the start point.

siniylev [52]

Answer:

Explanation:

The athlete ran a total distance of zero because they ran 100m forward then turned around so they went back to their starting position

6 0

3 years ago

A negative ion of charge -2e is located at the origin and a second negative ion of charge -3e is located nearby at x = 3.8 nm ,

Rus_ich [418]

Answer:

$\vec{F}_{21}=-5.63\times 10^{-11}N\\\\\vec{F}_{21}=\\$

Explanation:

Given that

$Q_1 = -2e\, C\\\\Q_2=-3e\,C\\\\x= 3.8 \times 10^{-9}\,m\\\\y= 3.2 \times 10^{-9}\,m\\\\r=\sqrt{x^2+y^2}\\\\r= 4.96\times 10^{-9} m\\$

As both charges are negative so there exist force of repulsion in direction as shown in figure.

$F_{12}=\frac{kQ_1Q_2}{r^2}\\\\F_{12}= \frac{(9\times 10^9)(6)(1.602\times 10^{-19})^2}{(4.96\times 10^{-9})^2}\\\\F_{12}=5.63\times 10^{-11}N$

Angle at which force F12 is acting is

$\theta=tan^{-1}\frac{3.2}{3.8}\\\\\theta=tan^{-1}\frac{y}{x}\\\\\theta= 40.1^o$

$F_{x}=F_{12}cos\theta\\\\F_{x}=(5.63\times 10^{-11})cos(40.1)\\\\F_{x}=4.306\times 10^{-11}N\\\\F_{y}=F_{12}sin\theta\\\\F_{y}=(5.63\times 10^{-11})sin(40.1)\\\\F_{y}=3.62\times 10^{-11}N\\\\$

$\vec{F}_{12}=\vec{F}_{x}+\vec{F}_y\\\\\vec{F}_{12}=4.30\times 10^{-11}\,\hat{i} + 3.62\times 10^{-11}\,\hat{j}\\\\\vec{F}_{12}=$

Force exerted on charge -2e is equal in magnitude to F12 but is in opposite direction

$F_{21}=-5.63\times 10^{-11}N$

$\vec{F}_{21}=$

7 0

4 years ago

When light reflects from a surface, there is a change in its

alexandr402 [8]

The answer is: none of the above.

Explanation:

When light reflects from a surface, the frequency, wavelength, and speed do not change. They remain the same.

6 0

3 years ago

How am I supposed to solve this?

RSB [31]

Answer:

4.02 km/hr

Explanation:

5 km/hr = 1.39 m/s

The swimmer's speed relative to the ground must have the same direction as line AC.

The vertical component of the velocity is:

uᵧ = us cos 45

uᵧ = √2/2 us

The horizontal component of the velocity is:

uₓ = 1.39 − us sin 45

uₓ = 1.39 − √2/2 us

Writing a proportion:

uₓ / uᵧ = 121 / 159

(1.39 − √2/2 us) / (√2/2 us) = 121 / 159

Cross multiply and solve:

159 (1.39 − √2/2 us) = 121 (√2/2 us)

220.8 − 79.5√2 us = 60.5√2 us

220.8 = 140√2 us

us = 1.115

The swimmer's speed is 1.115 m/s, or 4.02 km/hr.

7 0

3 years ago

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