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algol13
3 years ago
9

What is the approximate radius of pluto

Physics
2 answers:
krek1111 [17]3 years ago
3 0
1200 km to 2 s.f.

If you want to approximate it further then 1000 km.
Taya2010 [7]3 years ago
3 0
783.4 is the radius of pluto
 0.01303, 5.9724, 0.0022. Volume (1010<span> km</span>3<span>), 0.697, 108.321, 0.0064. Equatorial </span>radius<span> (km),</span>
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WILL MARK BRAINLIEST
Komok [63]

Answer:

yes every action as an equal and and opposite reaction. if you throw a tool then it will give you a reaction and you will move back.

Explanation:

According to Newton's third law of motion, when two bodies interact between them, appear equal forces and opposite senses in each of them.  

To understand it better:  

Each time a body or object exerts a force on a second body or object, it (the second body) will exert a force of equal magnitude but in the opposite direction on the first.  

So, if you as an astronaut in the described situation throw your tool  in the direction that you are traveling (in the opposite direction of the space station), according to Newton's third law, you will be automatically moving towards the station

6 0
2 years ago
A) explain how the data matches each line on the graph (give name and line letter for each)
stepladder [879]
The answer you have is right good job 12
3 0
2 years ago
What are the characteristics of the radiation emitted by a blackbody? According to Wien's Law, how many times hotter is an objec
jasenka [17]

Answer:

a) What are the characteristics of the radiation emitted by a blackbody?

The total emitted energy per unit of time and per unit of area depends in its temperature (Stefan-Boltzmann law).

The peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase (Wien’s displacement law).

The spectral density energy is related with the temperature and the wavelength (Planck’s law).

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wave length of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

Explanation:

A blackbody is an ideal body that absorbs all the thermal radiation that hits its surface, thus becoming an excellent emitter, as these bodies express themselves without light radiation, and therefore they look black.

The radiation of a blackbody depends only on its temperature, thus being independent of its shape, material and internal constitution.

If it is study the behavior of the total energy emitted from a blackbody at different temperatures, it can be seen how as the temperature increases the energy will also increase, this energy emitted by the blackbody is known as spectral radiance and the result of the behavior described previously is Stefan's law:

E = \sigma T^{4}  (1)

Where \sigma is the Stefan-Boltzmann constant and T is the temperature.

The Wien’s displacement law establish how the peak of emission of the spectrum will be displace to shorter wavelengths as the temperature increase (inversely proportional):

\lambda max = \frac{2.898x10^{-3} m. K}{T}   (2)

Planck’s law relate the temperature with the spectral energy density (shape) of the spectrum:

E_{\lambda} = {{8 \pi h c}\over{{\lambda}^5}{(e^{({hc}/{\lambda \kappa T})}-1)}}}  (3)

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wavelength of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

It is need it to known the temperature of both objects before doing the comparison. That can be done by means of the Wien’s displacement law.

Equation (2) can be rewrite in terms of T:

T = \frac{2.898x10^{-3} m. K}{\lambda max}   (4)

Case for the object with the blackbody emission spectrum peak in the blue:

Before replacing all the values in equation (4), \lambda max (450 nm) will be express in meters:

450 nm . \frac{1m}{1x10^{9} nm}  ⇒ 4.5x10^{-7}m

T = \frac{2.898x10^{-3} m. K}{4.5x10^{-7}m}

T = 6440 K

Case for the object with the blackbody emission spectrum peak in the red:

Following the same approach above:

700 nm . \frac{1m}{1x10^{9} nm}  ⇒ 7x10^{-7}m

T = \frac{2.898x10^{-3} m. K}{7x10^{-7}m}

T = 4140 K

Comparison:

\frac{6440 K}{4140 K} = 1.55

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

4 0
3 years ago
In a parallel circuit, if one connection is broken
nirvana33 [79]
C hope that helps hens
7 0
3 years ago
5.
iris [78.8K]

Answer:

the answer is a time your welcome

3 0
3 years ago
Read 2 more answers
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