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2 years ago

5

The energy that is radiated to Earth by the Sun includes visible light, infrared radiation, ultraviolet radiation, and other typ

es. These different forms of radiation are distinguished based on what property?

2 answers:

nasty-shy [4]2 years ago

7 0

Answer:

The energy that is radiated to Earth by the Sun includes visible light, infrared radiation, ultraviolet radiation./

Y_Kistochka [10]2 years ago

6 0

Answer:

The energy that is radiated to Earth by the Sun includes visible light, infrared radiation, ultraviolet radiation, and other types. These different forms of radiation are distinguished based on what property? Sunlight is a mixture of electromagnetic waves, ranging from infrared to ultraviolet.

Explanation:

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Which type of electromagnetic waves has highest frequency

vesna_86 [32]

Answer:

Gamma rays have the highest energies.

Explanation:

HOPE IT WILL HELP ^_^

6 0

3 years ago

Read 2 more answers

Problem 1: Two sources emit waves that are coherent, in phase, and have wavelengths of 26.0 m. Do the waves interfere constructi

Anton [14]

1) Destructive interference

The condition for constructive interference to occur is:

$\delta = m\lambda$ (1)

where

$\delta =|d_1 -d_2|$ is the path difference, with

$d_1$ is the distance of the point from the first source

$d_2$ is the distance of the point from the second source

m is an integer number

$\lambda$ is the wavelength

In this problem, we have

$d_1 = 78.0 m\\d_2 = 143 m\\\lambda=26.0 m$

So let's use eq.(1) to see if the resulting m is an integer

$\delta =|78.0 m-143 m|=65 m\\m=\frac{\delta }{\lambda}=\frac{65 m}{26.0 m}=2.5$

It is not an integer so constructive interference does not occur.

Let's now analyze the condition for destructive interference:

$\delta = (m+\frac{1}{2})\lambda$ (2)

If we apply the same procedure to eq.(2), we find

$m=\frac{\delta}{\lambda}-\frac{1}{2}=\frac{65.0 m}{26.0 m}-0.5=2$

which is an integer: so, this point is a point of destructive interference.

2) Constructive interference

In this case we have

$d_1 = 91.0 m\\d_2 =221.0 m$

So the path difference is

$\delta =|91.0 m-221.0 m|=130.0 m$

Using the condition for constructive interference:

$m=\frac{\delta }{\lambda}=\frac{130.0 m}{26.0 m}=5$

Which is an integer, so this is a point of constructive interference.

3) Destructive interference

In this case we have

$d_1 = 44.0 m\\d_2 =135.0 m$

So the path difference is

$\delta =|44.0 m-135.0 m|=91.0 m$

Using the condition for constructive interference:

$m=\frac{\delta }{\lambda}=\frac{91.0 m}{26.0 m}=3.5$

This is not an integer, so this is not a point of constructive interference.

So let's use now the condition for destructive interference:

$m=\frac{\delta}{\lambda}-\frac{1}{2}=\frac{91.0 m}{26.0 m}-0.5=3$

which is an integer: so, this point is a point of destructive interference.

3 0

3 years ago

Galileo started modern science because he started using ? Rather than speculation

Eva8 [605]

By using the telescope

4 0

3 years ago

A bowling ball rolls 33\,\text m33m33, start text, m, end text with an average speed of 2.5\,\dfrac{\text m}{\text s}2.5 s m 2

asambeis [7]

The ball rolled for 13.2 s

<h3>Further explanation</h3>

Speed is scalar and no direction

$\tt avg~speed=\dfrac{total~distance}{elapsed~time}=\dfrac{\Delta x}{\Delta t}$

A bowling ball rolls 33 m, with average speed = 2.5 m/s

So elapsed time :

$\tt t=\dfrac{33~m}{2.5`m/s}=13.2~s$

4 0

3 years ago

I WILL GIVE BRAINLIEST IF SOMEONE GETS THIS......

pav-90 [236]

Answer:

Explanation:

a)

Firstly to calculate the total mass of the can before the metal was lowered we need to add the mass of the eureka can and the mass of the water in the can. We don't know the mass of the water but we can easily find if we know the volume of the can. In order to calculate the volume we would have to multiply the area of the cross section by the height. So we do the following.

100 $cm^{2}$ x 10cm = 1000 $cm^{3}$

Now in order to find the mass that water has in this case we have to multiply the water's density by the volume, and so we get....

$\frac{1g}{cm^{3} }$ x 1000 $cm^{3}$ = 1000g or 1kg

Knowing this, we now can calculate the total mass of the can before the metal was lowered, by adding the mass of the water to the mass of the can. So we get....

1000g + 100g = 1100g or 1.1kg

b)

The volume of the water that over flowed will be equal to the volume of the metal piece (since when we add the metal piece, the metal piece will force out the same volume of water as itself, to understand this more deeply you can read the about "Archimedes principle"). Knowing this we just have to calculate the volume of the metal piece an that will be the answer. So this time in order to find volume we will have to divide the total mass of the metal piece by its density. So we get....

20g ÷ $\frac{8g}{cm^{3} }$ = 2.5 $cm^{3}$

c)

Now to find out the total mass of the can after the metal piece was lowered we would have to add the mass of the can itself, mass of the water inside the can, and the mass of the metal piece. We know the mass of the can, and the metal piece but we don't know the mass of the water because when we lowered the metal piece some of the water overflowed, and as a result the mass of the water changed. So now we just have to find the mass of the water in the can keeping in mind the fact that 2.5 $cm^{3}$ overflowed. So now we the same process as in number a) just with a few adjustments.

$\frac{1g}{cm^{3} }$ x (1000 $cm^{3}$ - 2.5 $cm^{3}$ ) = 997.5g

So now that we know the mass of the water in the can after we added the metal piece we can add all the three masses together (the mass of the can. the mass of the water, and the mass of the metal piece) and get the answer.

100g + 997.5g + 20g = 1117.5g or 1.1175kg

5 0

3 years ago