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irakobra [83]
3 years ago
14

Metals are good conductors of electricity because __________ . A) they are easily reduced and oxidised. B) they are easily ionis

ed C) their valence electrons are not localised D) they are ductile. E) they can be drawn into wires.
Physics
1 answer:
Advocard [28]3 years ago
3 0

Answer:

a

Explanation:

because they are easily reduced and oxidised

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Here are the answers:
1. Geosphere (though the term lithosphere is mostly used)
2. Both ice and wind (glaciers, and really strong winds)
3. Water
4. Its inertia (the Earth is constantly "falling" towards the Sun due to its gravitational pull, but its inertia helps the Earth from maintaining its orbit.)
5. The rotating Earth
6. one year
7. The equator
8. It depends on how much of the sunlit side of the Moon faces the Earth
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10. D<span>ifferences in how much the Moon and the Sun pull on different parts of Earth 
11. b. False
12. a. True

Hope my answers have come to your help.</span>
4 0
3 years ago
Is calculating the change of velocity the same as calculating acceleration? ​
koban [17]

Answer:

Yes! Thinking about it graphically a position vs time graph models meters per second in most cases, making every point on the line have the units m/s. If we want the find the slope we are finding the change between each point and those units would change to m/s/s or m/s^2 giving us the same units for acceleration. Simply put, slope of a velocity graph gives us acceleration.

Explanation:

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2 years ago
DEFINE the term free fall
PtichkaEL [24]

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when you get to fall for free

Explanation:

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3 years ago
Read 2 more answers
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.

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3 years ago
The light from the sun has higher frequencies from one side of the sun than from the other side. What does that tell you about t
morpeh [17]

If the light from the sun has higher frequencies from one side of the sun than from the other side, it is proof that the Sun is rotating.

Doppler effect states that, if a person is standing still and a source ( sound / light ) is moving towards him, the frequency of the wave emitted from the object will increase and if the source ( sound / light ) is away from him, the frequency of the wave emitted from the object will decrease.

So, if the light from the sun has higher frequencies from one side of the sun than from the other side, it means that the Sun is rotating. The higher frequencies points are the points that rotating towards Earth and lower frequencies points are the points that rotating away from Earth.

Therefore, if the light from the sun has higher frequencies from one side of the sun than from the other side, it is proof that the Sun is rotating.

To know more about Doppler Effect

brainly.com/question/15318474

#SPJ1

8 0
1 year ago
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