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2 years ago

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Question 20 Unsaved The graph below represents changes in water. What process is occurring during section "B" of the graph? Ques

tion 20 options: vaporization condensation freezing melting

2 answers:

juin [17]2 years ago

5 0

melting ... happpens at constant temp

FinnZ [79.3K]2 years ago

4 0

melting ... happpens at constant temp

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Which of the following is fact-based science rather than part of a personal belief system?

marusya05 [52]

Since there are no choices, then this question calls for open-ended answers. Facts-based science must have proven underlying laws that support inferences such as Coulomb's Law, Kinetic Theory of Matter and many more. On the other hand, examples of science that focus on personal belief is philosophy. This depends on the perspective of known philosophers. An example would be Sigmund Freud who proposed the theory of 3 personalities. Although it is more on personal beliefs, this is used as a foundation in the study of psychology.

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2 years ago

Label the following descriptions with (M) for Mass or (W) for Weight. ____measured with a spring scale ____measured in Newtons _

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1. mass
2. weight
3. weight
4. weight
5. mass
6. mass

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2 years ago

Of the following transitions in the Bohr hydrogen atom, the ________ transition results in the absorption of the highest-energy

8_murik_8 [283]

Answer:

The <em><u>n = 2 → n = 3</u></em> transition results in the absorption of the highest-energy photon.

Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

Formula used for the radius of the $n^{th}$ orbit will be,

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

Here: Z = 1 (hydrogen atom)

Energy of the first orbit in H atom .

$E_1=-13.6\times \frac{Z^2}{1^2} eV=-13.6 eV$

Energy of the second orbit in H atom .

$E_2=-13.6\times \frac{Z^2}{(2)^2} eV=-3.40 eV$

Energy of the third orbit in H atom .

$E_3=-13.6\times \frac{Z^2}{(9)^2} eV=-1.51 eV$

Energy of the fifth orbit in H atom .

$E_5=-13.6\times \frac{Z^2}{(2)^2} eV=-0.544 eV$

Energy of the sixth orbit in H atom .

$E_6=-13.6\times \frac{Z^2}{(2)^2} eV=-0.378 eV$

Energy of the seventh orbit in H atom .

$E_7=-13.6\times \frac{Z^2}{(2)^2} eV=-278 eV$

During an absorption of energy electron jumps from lower state to higher state.So, absorption will take place in :

1) n = 2 → n = 3

2) n= 5 → n = 6

Energy absorbed when: n = 2 → n = 3

$E=E_3-E_2$

$E=(-1.51 eV) -(-3.40 eV)=1.89 eV$

Energy absorbed when: n = 5 → n = 6

$E'=E_6-E_5$

$E'=(-0.378 eV)-(-0.544 eV) =0.166 eV$

1.89 eV > 0.166 eV

E> E'

So,the n = 2 → n = 3 transition results in the absorption of the highest-energy photon.

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2 years ago

A 5,000 kg satellite is orbiting the Earth in a circular path. The height of the satellite above the surface of the Earth is 800

Arada [10]

Explanation:

The given data is as follows.

m = 5000 kg, h = 800 km = $0.8 \times 10^{6} m$

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$M_{e} = 5.98 \times 10^{24}$ kg, G = $6.67 \times 10^{-11} Nm^{2}/kg^{2}$

As we know that,

$\frac{mv^{2}}{r} = \frac{GmM_{e}}{r^{2}}$

v = $\sqrt{\frac{GM_{e}}{r^{2}}}$

And, it is known that formula to calculate angular velocity is as follows.

$\omega = \frac{v}{r} = \sqrt{\frac{GM_{e}}{r^{3}}}$

v = $\sqrt{\frac{GM_{e}}{r^{3}}}$

= $\sqrt{\frac{6.67 \times 10^{-11} Nm^{2}/kg^{2} \times 5.98 \times 10^{-24} kg^{-2}}{(7.17 \times 10^{6} m)^{3}}}$

= $1.0402 \times 10^{-3} rad/s$

Thus, we can conclude that speed of the satellite is $1.0402 \times 10^{-3} rad/s$ .

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On a nice winter day at the South Pole, the temperature rises to −54°F. What is the approximate temperature in degrees Celsius?

Evgesh-ka [11]

Answer:

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2 years ago

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