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aleksklad [387]
3 years ago
9

Please help me ,create a slogan that promotes the role of internet in education

Physics
1 answer:
aleksklad [387]3 years ago
7 0

Answer:

¨Facts you didn´t know¨ or ¨unknown facts¨

Explanation:

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gravity is a force between any two objects with mass. why doesn't a person feel a gravitational force between him/herself and an
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c)the gravitational forces of people is so small it is overshadowed by that of earth.

Explanation:

The gravitational force between two objects is given by:

F=G\frac{m_1 m_2}{r^2}

where

G is the gravitational constant

m1 and m2 are the masses of the two objects

r is the distance between the two objects

From the formula, we see that the gravitational force depends on the masses of the objects: since the mass of the Earth (5.97\cdot 10^{24} kg is much much larger than the average mass of one person (80-100 kg), then the gravitational force exerted by the Earth on a person is also much much larger than the gravitational force between two people.

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Please help on all i cant figure out
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In which situation is the speed of the car constant while its velocity is changing?
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Answer:

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Explanation:

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2 years ago
A practical rule is that a radioactive nuclide is essentially gone after 10 half-lives. What percentage of the original radioact
ArbitrLikvidat [17]

Answer:

  • 0.09 % of the original radioactive nucllde its left after 10 half-lives
  • It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

Explanation:

The equation for radioactive decay its:

N ( t) \ = \ N_0 \ e^{ \ -  \frac{t}{\tau}},

where N(t) its quantity of material at time t, N_0 its the initial quantity of material and \tau its the mean lifetime of the radioactive element.

The half-life t_{\frac{1}{2}} its the time at which the quantity of material its the half of the initial value, so, we can find:

N (t_{\frac{1}{2} }) \ = \ N_0 \ e^{ \ -  \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}

so:

\ N_0 \ e^{ \ -  \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}

e^{ \ -  \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{1}{2}

-  \frac{t_{\frac{1}{2}}}{\tau}} \ = - \ ln( 2 )

t_{\frac{1}{2}}\ = \tau ln( 2 )

So, after 10 half-lives, we got:

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ -  \frac{10 \  t_{\frac{1}{2}}}{\tau}}

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ -  \frac{10 \  \tau \ ln( 2 ) }{\tau}}

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ -  10 \  \ ln( 2 ) }

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ * \ 9.76 * 10^{-4}

So, we got that a 0.09 % of the original radioactive nucllde its left.

Putonioum-239 has a half-life of 24,110 years. So, 10 half-life will take to pass

10 \ * \ 24,110 \ years \ = \ 241,100 \ years

It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

7 0
3 years ago
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