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Allushta [10]
3 years ago
14

From the illustration of the potassium atom, fill in the periodic table selection.

Physics
1 answer:
stich3 [128]3 years ago
3 0

Answer:

2 in shell 1; 8 in shell 2; 8 in shell 3 and 1 in shell 4 ; 19 total protons and electrons

Explanation: Potassium is in Group 1 Period 4 and will have 1 valence electron that is 4s1 electron configuration.

Atomic mass =

39.10  = 19 protons + 20 neutrons

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Every action has an equal or opposite reaction.
You weigh 60kg 
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What are the characters associated with light as a wave​
Liono4ka [1.6K]

Answer:

Interference of light

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6 0
3 years ago
An object of mass 3 kg, moving with and sticks to an object of mass 2.00 kg with an initial velocity of 3.00 m/s. Find the final
Aleonysh [2.5K]

Answer:

1.8 m/s

Explanation:

momentum = mass × velocity

initial momentum = m1v1+m2v2

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let combined velocity be V

HENCE  

final momentum = total mass × velocity

                            = (3+2) × V = 5V

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final momentum = initial momentum

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8 0
3 years ago
which pole of a compass needle points north ? please give simple answer with little explaining if any :)
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50POINTS! Find the orbital speed of a satellite in a circular orbit 1700km above the surface of the Earth. M_earth 5.97e24kg, r_
ivolga24 [154]
<h2>Answer: 7020.117 m/s</h2>

Explanation:

The velocity of a satellite describing a circular orbit is<u> constant</u> and defined by the following expression:  

V=\sqrt{G\frac{M}{R}} (1)  

Where:  

G=6.674({10}^{-11})\frac{N{m}^{2}}{{kg}^{2}} is the gravity constant

M_{Earth}=5.97{10}^{24}kg the mass of the massive body around which the satellite is orbiting, in this case, the Earth .

R=r_{Earth}+h=8080000m the radius of the orbit (measured from the center of the planet to the satellite).  

This means the radius of the orbit is equal to <u>the sum</u> of the average radius of the Earth r_{Earth} and the altitude of the satellite above the Earth's surface h.

Note this orbital speed, as well as orbital period, does not depend on the mass of the satellite. It depends on the mass of the massive body (the Earth).

Now, rewriting equation (1) with the known values:

V=\sqrt{(6.674({10}^{-11})\frac{N{m}^{2}}{{kg}^{2}})\frac{5.97{10}^{24}kg}{8080000m}}

V=7020.117\frac{m}{s}  

6 0
3 years ago
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