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Lorico [155]
2 years ago
12

Which best describes the current atomic theory? (1 point) a Since it is only a theory it should not be used in practice. b It ha

s not been tested enough to be useful for developing technology. c It is the most powerful explanation scientists have to offer at this time. d There is very little chance it will be changed in the future.
Physics
2 answers:
Ahat [919]2 years ago
8 0

Answer:scientist try to fix it

Explanation:

gayaneshka [121]2 years ago
6 0

Answer:

Hello there buddy

Your answer is

d There is very little chance it will be changed in the future.

Hope it’s helps you buddy

I’m really sorry if my answer is wrong

Explanation:

~ZaynTheEpicFoxy

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The maximum wavelength that an electromagnetic wave can have and still eject electrons from a metal surface is 459 nm. What is t
spin [16.1K]

Answer:

2.7067 eV

Explanation:

h = Planck's constant = 6.626\times 10^{-34}\ m^2kg/s

c = Speed of light = 3\times 10^8\ m/s

\lambda_0 = Threshold wavelength = 459 nm

Work function is given by

W_0=\frac{hc}{\lambda_0}\\\Rightarrow W_0=\frac{6.626\times 10^{-34}\times 3\times 10^8}{459\times 10^{-9}}\\\Rightarrow W_0=4.33072\times 10^{-19}\ J

Converting to eV

1\ J=\frac{1}{1.6\times 10^{-19}}\ eV

4.33072\times 10^{-19}\ J=4.33072\times 10^{-19}\times \frac{1}{1.6\times 10^{-19}}\ eV=2.7067\ eV

The work function W0 of this metal is 2.7067 eV

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3 years ago
Where do you think the government should put the greatest support: solar energy, wind energy, clean coal, oil exploration, gas e
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The government should put it support in a combination of sources, as no source in the present can fully provide all energy requirements.
7 0
3 years ago
Read 2 more answers
How does an atom of bromine-79 become a bromide ion with a -1 charge?
Tanzania [10]

a the atom loses 1 proton to have a total of 34

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3 years ago
How do we use information from earths climate change data in order to engineer soulutions that reduce carbon emissions and impac
masya89 [10]

Answer:

That is a very broad question. One thing that does not seem to be considered is the depletion of the ozone layer at high altitudes.

In the 1960's chlorofluorcarbons (CFC,s) became popular as refrigerants, spray can propellants, etc.  In January 1989 the Montreal Protocol was passed which has greatly reduced the use of these substances. However, it may be several decades before the ozone layer can be replaced and again absorb harmful ulraviolet rays that may be partly responsible for the increase in global warming.  

(One chlorine atom at high altitudes can be responsible for the destruction of 100,000 molecules of ozone - catalytic reaction)

7 0
2 years ago
During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun.
lord [1]

Answer:

(a) F_{sm} = 4.327\times 10^{20}\ N

(b) F_{em} = 1.983\times 10^{20}\ N

(c) F_{se} = 3.521\times 10^{20}\ N

Solution:

As per the question:

Mass of Earth, M_{e} = 5.972\times 10^{24}\ kg

Mass of Moon, M_{m} = 7.34\times 10^{22}\ kg

Mass of Sun, M_{s} = 1.989\times 10^{30}\ kg

Distance between the earth and the moon, R_{em} = 3.84\times 10^{8}\ m

Distance between the earth and the sun, R_{es} = 1.5\times 10^{11}\ m

Distance between the sun and the moon, R_{sm} =  1.5\times 10^{11}\ m

Now,

We know that the gravitational force between two bodies of mass m and m' separated by a distance 'r' is given y:

F_{G} = \frac{Gmm'_{2}}{r^{2}}                             (1)

Now,

(a) The force exerted by the Sun on the Moon is given by eqn (1):

F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}

F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}

F_{sm} = 4.327\times 10^{20}\ N

(b) The force exerted by the Earth on the Moon is given by eqn (1):

F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}

F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}

F_{em} = 1.983\times 10^{20}\ N

(c) The force exerted by the Sun on the Earth is given by eqn (1):

F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}

F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}

F_{se} = 3.521\times 10^{20}\ N

7 0
3 years ago
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