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abruzzese [7]
3 years ago
13

How do you determine the number of electrons in the shells

Physics
1 answer:
snow_tiger [21]3 years ago
4 0

Periodic table

Find your element

Locate the element atomic number

Determine the number of electrons

Look for the atomic mass of the element

subtract the atomic number from the atomic number  

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Whenever two Apollo astronauts were on the surface of the Moon, a third astronaut orbited the Moon. Assume the orbit to be circu
inysia [295]

Answer:

v  =  1,582 \ \frac{m}{s}

Explanation:

We know that for circular motion the centripetal acceleration a_c is:

a_c = \frac{v^2}{r}

where v is the speed and r is the radius.

The centripetal acceleration for the astronaut must be the gravitational acceleration due to the gravity, as there are no other force. So

a_c = 1.27 \frac{m}{s^2}.

The radius of the orbit must be the radius of the Moon, plus the 270 km above the surface

r = 1.7 * 10^6 \ m + 270  \ km

r = 1.7 * 10^6 \ m + 270 * 10^3 \ m

r = 1.7 * 10^6 \ m + 0.270 * 10^6 \ m

r = 1.97 * 10^6 \ m

We can obtain the speed as:

v^2  = a_c r

v  = \sqrt{a_c r}

v  = \sqrt{1.27 \frac{m}{s^2} * 1.97 * 10^6 \ m}

v  = \sqrt{ 2.509 \ 10^6 \ \frac{m^2}{s^2}}

v  =  1.582 \ 10^3 \ \frac{m}{s}

v  =  1,582 \ \frac{m}{s}

And this is the orbital speed.

7 0
3 years ago
What are basic physical quantities and it's measurement units? Make a chart.
jolli1 [7]

Displacement/distance metres
Time seconds
Force Newtons
Energy Joules
Voltage Volts
Current intensity Amperes
Resistance Ohms
Light intensity Candella
Pressure Pascals
Charge Coulombs

8 0
3 years ago
Two electrons in a vacuum exert force of F = 3.8E-09 N on each other. They are then moved such that they are separated by x = 8.
iren [92.7K]

Answer:

F_n = 5.65E-11 N

d =  1.20682E-31 m

Explanation:

F = 3.8E-09 N

where

m = Mass of electron = 9.109E−31 kilograms

G = Gravitational constant = 6.67E-11 m³/kgs²

x = Distance between them

F=G\frac{m^2}{x^2}\\\Rightarrow 3.8E-09=G\frac{m^2}{x^2}

For F_n

F_n=G\frac{m^2}{x^2}\\\Rightarrow F_n=G\frac{m^2}{(8.2x)^2}\\\Rightarrow F_n=G\frac{m^2}{67.24x^2}

Dividing the above equations we get

\frac{F}{F_n}=\frac{G\frac{m^2}{x^2}}{G\frac{m^2}{67.24x^2}}\\\Rightarrow \frac{F}{F_n}=67.24\\\Rightarrow F_n=\frac{F}{67.24}\\\Rightarrow F_n=\frac{3.8E-09}{67.24}\\\Rightarrow F_n=5.65E-11\ N

F_n = 5.65E-11 N

F=G\frac{m^2}{x^2}\\\Rightarrow x=\sqrt{\frac{Gm^2}{F}}\\\Rightarrow x=\sqrt{\frac{G}{F}}m\\\Rightarrow x=\sqrt{\frac{6.67E-11}{3.8E-09}}9.109E-31\\\Rightarrow x=1.20682E-31\ m

d =  1.20682E-31 m

8 0
3 years ago
Suppose the coefficient of static friction between the road and the tires on a car is 0.638 and the car has no negative lift. Wh
larisa [96]

Answer:

12.6332454263 m/s

Explanation:

m = Mass of car

v = Velocity of the car

\mu = Coefficient of static friction = 0.638

g = Acceleration due to gravity = 9.81 m/s²

r = Radius of turn = 25.5 m

When the car is on the verge of sliding we have the force equation

\dfrac{mv^2}{r}=\mu mg\\\Rightarrow v=\sqrt{\mu gr}\\\Rightarrow v=\sqrt{0.638\times 9.81\times 25.5}\\\Rightarrow v=12.6332454263\ m/s

The speed of the car that will put it on the verge of sliding is 12.6332454263 m/s

4 0
3 years ago
I NEED PHYSICS HELP ASAP! DO NOT GUESS! I WILL GIE BRAINLEIST IF RIGHT! THANKS! :D
ale4655 [162]
<span>The repelling of the support magnet decreases friction. is the answer you're looking for .  :)

hope i helped - beanz</span>
6 0
3 years ago
Read 2 more answers
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