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1 year ago

The diagram below shows how sunlight is radiated back into the Earth's atmosphere as heat (infrared radiation).

Physics

1 answer:

Juliette [100K]1 year ago

6 0

The diagram that shows how sunlight is radiated back into the Earth’s atmosphere as heat(infrared radiation) is of the Greenhouse effect.

The greenhouse effect is a natural phenomenon that occurs on Earth and heats up our atmosphere.

The name of this effect is derived from a greenhouse, which is used to grow crops in a special environment.

A greenhouse is basically a house that is made up of glass. The glass allows heat to enter the house but it does not allow the heat to escape to the outside thus keeping the greenhouse warm.

Some gases that are responsible for this effect are known as greenhouse gases. These gases are carbon dioxide, methane, chloro- fluoro carbons, etc.

These gases trap the heat of the sun and raise the temperature of Earth thus causing global warming.

Thus, the diagram that shows how sunlight is radiated back into the Earth’s atmosphere as heat(infrared radiation) is of the Greenhouse effect.

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Time period from one noon to the next

matrenka [14]

That's the period of time known as one solar "day". We subdivide it into 24 slices which we call "hours". Using this system of time units, the day is about 4 minutes longer than one complete axial rotation of the Earth.

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3 years ago

What is a plane mirror? state the characteristics of the image formed by a plane mirror

Nadusha1986 [10]

A plane mirror always forms a virtual image. the image and the object are the same distance from a flat mirror, the image size is the same as the object, and the image is upright!

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2 years ago

3. The center of mass (or center of gravity) of a two-particle system is at the origin. One particle

nika2105 [10]

Answer:

B) (-2.0 m, 0.0 m)

Explanation:

Given:

Mass of particle 1 is, $m_1=2.0\ kg$

Mass of particle 2 is, $m_2=3.0\ kg$

Position of center of mass is, $(x_{cm},y_{cm})=(0,0)$

Position of particle 1 is, $(x_1,y_1)=(3.0\ m,0.0\ m)$

Position of particle 2 is, $(x_2,y_2)=(?\ m,?\ m)$

We know that, the x-coordinate of center of mass of two particles is given as:

$x_{cm}=\frac{m_1x_1+m_2x_2}{m_1+m_2}$

Plug in the values given.

$0=\frac{(2.0\times 3)+(3.0\times x_2)}{2.0+3.0}\\\\0=6+3x_2\\\\3x_2=-6\\\\x_2=\frac{-6}{3}=-2.0\ m$

We know that, the y-coordinate of center of mass of two particles is given as:

$y_{cm}=\frac{m_1y_1+m_2y_2}{m_1+m_2}$

Plug in the values given.

$0=\frac{(2.0\times 0)+(3.0\times y_2)}{2.0+3.0}\\\\0=0+3y_2\\\\3y_2=0\\\\y_2=\frac{0}{3}=0.0\ m$

Therefore, the position of particle 2 of mass 3.0 kg is (-2.0 m, 0.0 m).

So, option (B) is correct.

8 0

3 years ago

A large cylindrical tank contains 0.750 cubic meters of nitrogen gas at 27 degrees celsius and 1.5 e5 pa absolute pressure. the

k0ka [10]

<span>3.36x10^5 Pascals The ideal gas law is PV=nRT where P = Pressure V = Volume n = number of moles of gas particles R = Ideal gas constant T = Absolute temperature Since n and R will remain constant, let's divide both sides of the equation by T, getting PV=nRT PV/T=nR Since the initial value of PV/T will be equal to the final value of PV/T let's set them equal to each other with the equation P1V1/T1 = P2V2/T2 where P1, V1, T1 = Initial pressure, volume, temperature P2, V2, T2 = Final pressure, volume, temperature Now convert the temperatures to absolute temperature by adding 273.15 to both of them. T1 = 27 + 273.15 = 300.15 T2 = 157 + 273.15 = 430.15 Substitute the known values into the equation 1.5E5*0.75/300.15 = P2*0.48/430.15 And solve for P2 1.5E5*0.75/300.15 = P2*0.48/430.15 430.15 * 1.5E5*0.75/300.15 = P2*0.48 64522500*0.75/300.15 = P2*0.48 48391875/300.15 = P2*0.48 161225.6372 = P2*0.48 161225.6372/0.48 = P2 335886.7441 = P2 Rounding to 3 significant figures gives 3.36x10^5 Pascals. (technically, I should round to 2 significant figures for the result of 3.4x10^5 Pascals, but given the precision of the volumes, I suspect that the extra 0 in the initial pressure was accidentally omitted. It should have been 1.50e5 instead of 1.5e5).</span>

8 0

3 years ago

Determine the components of reaction at the fixed support

denis-greek [22]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The components of reaction at the fixed support are

$A_{(x)} = 400 \ N$ , $A_{(y)} = -500 \ N$ , $A_{(z)} = 600 \ N$ , $M_x = 1225 \ N\cdot m$ , $M_y = 750 \ N\cdot m$ , $M_z = 0 \ N\cdot m$