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lana [24]
3 years ago
14

The chart below shows the average surface temperature or temperature range for each of the eight planets.

Physics
2 answers:
Vaselesa [24]3 years ago
4 0

Answer:

A) Earth and the other inner planets have higher average surface temperatures than the outer planets.

Explanation:

the earth and the other inner planets have higher average surface temperatures than the outer planets.

The reason for this response is due to the distance between the sun and the respective planet, the source of energy generation is the sun and the only way in which the temperature increase of each planet is guaranteed is by radiation, the further away a planet is from its star, its temperature will decrease. Although it is also important to highlight the atmospheric composition of the planet if this planet in its stratosphere has high density clouds that do not allow the entry of solar radiation, the temperature of the planet's surface will not increase, independent of the distance from the sun, but these are more complex cases where specialists in that area enter to perform a study in detail.

MakcuM [25]3 years ago
4 0

Answer:

Earth and the other inner planets have higher average surface temperatures than the outer planets.

Explanation:

The chart shows that the temperatures of the inner planets (Mercury, Venus, Earth, and Mars) range from 100 K to 735 K. The temperatures of the outer planets range from 55 K to 165 K.

Thus, a good conclusion for the data would be that Earth and the other inner planets have higher average surface temperatures than the outer planets.

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Which of the following can penetrate the deepest (Please explain)
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Answer: 3MeV electron

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m_e={9.1\times 10^{-31}      m_α=4\times m_e    m_a={9.1\times 10^{-31}

m_p=1.67\times 10^{-27}

(a)  K.E. Energy of electron =\frac{1}{2}\times{m_e}\times{v_{e} ^{2}}=3MeV

v_{e} ^{2}=\frac{2\times3\times1.6\times10^{-19}\times10^{6}  }{9.1\times 10^{-31} }=1.05\times10^{18}

v_e=\sqrt{1.05\times10^{18} } = 1.025\times10^{9}\frac{m}{s}

(b) K.E. Energy of alpha particle =\frac{1}{2}\times{m_\alpha}\times{v_{\alpha} ^{2}}=10MeV

v_{\alpha} ^{2}= \frac{2\times10\times1.6\times10^{-19}\times10^{6}  }{4\times9.1\times 10^{-31} }=0.88\times10^{18}

v_\alpha=\sqrt{0.88\times10^{18} } =.94\times10^{9}\frac{m}{s}

(c) K.E. Energy of auger particle =\frac{1}{2}\times{m_a}\times{v_{a} ^{2}}=0.1MeV

v_{a} ^{2}=\frac{2\times0.1\times1.6\times10^{-19}\times10^{6}  }{9.1\times 10^{-31} }=0.035\times10^{18}

v_a=\sqrt{0.035\times10^{18} } =.19\times10^{9}\frac{m}{s}

(d)  K.E. Energy of proton particle =\frac{1}{2}\times{m_p}\times{v_{p} ^{2}}=400keV

v_{p} ^{2}=\frac{2\times400\times1.6\times10^{-19}\times10^{3}  }{1.67\times 10^{-27} }=0.766\times10^{14}

v_p=\sqrt{0.766\times10^{14} } =0.875\times10^{7}[tex]\frac{m}{s}

from (a),(b),(c),and (d) we can clearly say that the velocity of the electron is more so the penetration of the electron will be deepest.

6 0
3 years ago
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