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2 years ago

Which of the following is true of greenhouse gases?

Physics

1 answer:

Sedaia [141]2 years ago

4 0

Answer:

A. They trap energy in the atmosphere.

Explanation:

Greenhouse gases are defined as gases that absorb and emit radiation within the infrared range. These gases allow the sun’s rays to pass through the ozone layer and warm the earth, but prevent this warmth from escaping atmosphere.

Greenhouse gases cause the greenhouse effect on planets. Examples of greenhouse gases in Earth's atmosphere are water vapor, carbon dioxide, methane, ozone etc.

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The lightbulb is an example of

AlexFokin [52]

A object that has been reinvented so it is more energy efficient

4 0

3 years ago

Find the volume of the irregular object a marble

Lostsunrise [7]

By using Displacement method we can find volume of irregular object like marble.

<h3>What is displacement method?</h3>

In displacement method,

First , we measuring the volume of water displaced by an object which tell us the volume of the object.

Secondly We can use the physical balance to determine its mass.

Lastly , calculate the density by dividing the mass by the volume.

Procedure to find the Volume of irregular object i.e. marble

step 1 - Fill the graduated cylinder about half full and measure the initial volume of water.

step 2 - Drop the marble in the graduated cylinder.

step 3 -Now measure the final level of water.

Step 4 - Subtract both the values.

Here , comes the Volume of Irregular Object.

For more volume related question visit here:

brainly.com/question/1578538

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7 0

1 year ago

which statement best describes what would happen if the number of wore coils in a electromagnavneg were increased

jekas [21]

Where are the following choices

4 0

3 years ago

With detailed explaniation

belka [17]

Ø=37°
Initial velocity=u=20m/s
g=10m/s²

$\\ \rm\Rrightarrow H_{max}=\dfrac{u^2sin^2\theta}{2g}$

$\\ \rm\Rrightarrow H_{max}=\dfrac{20^2(sin37)^2}{2(10)}$

$\\ \rm\Rrightarrow H_{max}=\dfrac{400sin^237}{20}$

$\\ \rm\Rrightarrow H_{max}=20sin^237$

$\\ \rm\Rrightarrow H_{max}=7.2m$

$\\ \rm\Rrightarrow R=\dfrac{u^2sin2\theta}{g}$

$\\ \rm\Rrightarrow R=\dfrac{20^2sin74}{10}$

$\\ \rm\Rrightarrow R=40sin74$

$\\ \rm\Rrightarrow R=38.5m$

$\\ \rm\Rrightarrow T=\dfrac{2usin\theta}{g}$

$\\ \rm\Rrightarrow T=\dfrac{2(20)sin37}{10}$

$\\ \rm\Rrightarrow T=4sin37$

$\\ \rm\Rrightarrow T=2.4s$

Now

$\\ \rm\Rrightarrow v=u-gt$

$\\ \rm\Rrightarrow v=20-10(2.4)$

$\\ \rm\Rrightarrow v=20-24$

$\\ \rm\Rrightarrow v=-4m/s$

4 0

2 years ago

Suppose that a planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average o

Musya8 [376]

Explanation:

It is given that,

A planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average orbit radius of Mercury.

Mass of the Sun, $M=1.99\times 10^{30}\ kg$

Radius of Mercury's orbit, $r=5.79\times 10^{10}\ m$

Radius of discovered planet, $R=\dfrac{2}{3}r$

$R=\dfrac{2}{3}\times 5.79\times 10^{10}\ m=3.86\times 10^{10}\ m$

Let T is the orbital period of such a planet. Using Kepler's third law of planetary motion as :

$T^2\propto R^3$

$T^2=\dfrac{4\pi^2R^3}{GM}$

$T^2=\dfrac{4\pi^2\times (3.86\times 10^{10})^3}{6.67\times 10^{-11}\times 1.99\times 10^{30}}$

$T=\sqrt{1.71\times 10^{13}}$

T = 4135214.625 s

T = 47.86 days

So, the orbital period of such a planet is 47.86 days. Hence, this is the required solution.

6 0

2 years ago