In which direction would a positive charge move if it were placed in between the positive and negative charge

During which change of state do atoms lose energy? sublimation freezing boiling melting

Lorico [155]

Answer:

B) freezing

Explanation:

I just took the test on Edgenuity

3 0

3 years ago

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archaeologists use radioactive decay of an isotope of the element because it has a fairly short half-life and is found in anythi

natulia [17]

Archaeologists use radioactive decay of an isotope of the element <u>carbon.</u> Because it has a fairly short half-life, and is found in anything that was once living.

<h3>What is radioactivity?</h3>

The act of producing radiation spontaneously is known as radioactivity. This is accomplished by an unstable atomic nucleus that want to give up some energy in order to move to a more stable form.

Archaeologists use radioactive decay of an isotope of the element <u>carbon.</u> Because it has a fairly short half-life, and is found in anything that was once living.

Hence, <u>carbon</u> is used the correct answer for the blank.

To learn more about the radioactivity, refer to the link;

brainly.com/question/1770619

#SPJ1

7 0

2 years ago

10. A triply ionized beryllium atom is in the ground state. It absorbs energy and makes a transition to the n = 5 excited state.

Xelga [282]

Answer:

$\lambda=1282\ nm$

Explanation:

$E_n=-2.179\times 10^{-18}\times \frac{1}{n^2}\ Joules$

For transitions:

$Energy\ Difference,\ \Delta E= E_f-E_i =-2.179\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

$\Delta E=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

Also, $\Delta E=\frac {h\times c}{\lambda}$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

So,

$\frac {h\times c}{\lambda}=2.179\times 10^{-18}(|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)\ J$

$\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)}\ m$

So,

$\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)}\ m$

Given, $n_i=5\ and\ n_f=3$

$\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (\frac{1}{5^2} - \dfrac{1}{3^2})}\ m$

$\lambda=\frac{10^{-26}\times \:19.878}{10^{-18}\times \:2.179\left(|\frac{1}{25}-\frac{1}{9}\right)|}\ m$

$\lambda=\frac{19.878}{10^8\times \:2.179\left(|-\frac{16}{225}\right|)}\ m$

$\lambda= 1.2828\times10^{-6}$

1 m = 10⁻⁹ nm

$\lambda=1282\ nm$

6 0

3 years ago

Which pm the following are likely to form a covalent bond

Alecsey [184]

Answer:

magneisuim and gold

Explanation:

8 0

4 years ago

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100 points! please help!!!

IRINA_888 [86]

Answer:

Explanation:

A plane flies due north (90° from east) with a velocity of 100 km/h for 2 hours.

With no wind, it will be 100*2 = 200 km north of its starting point.

But a steady wind blows southeast at 30 km/h at an angle of 315° from due east.

So the wind itself will blow the plane 30*2 = 60km at an angle of 315° from due east.

That is the same as 60*cos315° = 42.43km due east and 60*sin315° = -42.43km north.

Combining, the plane is at 42.43km due east and 200-42.43 = 157.57km due north from its starting point.

7 0

3 years ago

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