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harina [27]
3 years ago
13

A Cassegrain telescope

Physics
1 answer:
yuradex [85]3 years ago
3 0

The answer is D.

Hope this helps and have a good day :D

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Using the periodic table entry below, match the phrases with their corresponding values.
Anika [276]

<u>Answer:</u>

For 1: Atomic number :  79

For 2. Atomic mass : 197

For 3. Number of orbits  : 6

For 4. Number of neutrons : 118

For 5. Number of electrons in n=3 : 18

For 6. Number of valence electrons : 1

<u>Explanation:</u>

  • Atomic number is defined as the number of protons or electrons present in an atom. Here 79 corresponds to the Atomic Number.
  • Atomic mass is defined as the Sum of neutrons and protons present in an atom. Here, 197 corresponds to the Atomic Mass.
  • Atomic Mass = Number of protons + Number of neutrons

Here, Number of protons = 79

So, 197=\text{Number of Neutrons}+79\\\\\text{Number of neutrons}=118

  • Valence electronic configuration of  the atom having atomic number as 79 is : [Xe]5d^{10}6s^1

Number of orbit corresponds the higher value of 'n' which is the principle quantum number. Here, n = 6 so, number of orbits are also 6.

  • In orbit having n = 3, there are 3 sub-shells: s, p and d. The number of electrons that can be occupied in s-subshell = 2

The number of electrons that can be occupied in p-subshell = 6

The number of electrons that can be occupied in d-subshell = 10

Total number of electrons that can be occupied in  orbit having n = 3 are (2 + 6 + 10) = 18 electrons.

  • From the electronic configuration, it is visible that valence electrons in atom having atomic number 79 is 1.
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c

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A sound whispered at one focus of a whispering gallery can be heard at the other focus. Two people are standing in a whispering
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In introductory physics laboratories, a typical Cavendish balance for measuring the gravitational constant G uses lead spheres o
SVETLANKA909090 [29]

Answer:

F = 7.7*10^{-10}N

Explanation:

You need to be careful with units for this problem. The force will be:

F =\frac{K*m1*m2}{d^2}

F=\frac{6.67259 * 10^{-11}*1.56*21.1*10^{-3}}{(5.34*10^{-2})^2}

F=7.7*10^{-10}N

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Answer: the atmosphere would still be in motion with the Earth's original 1100 mile per hour rotation speed at the equator.

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