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4vir4ik [10]
3 years ago
6

How many more electrons can fit in the valence shell of a fluorine atom?

Physics
1 answer:
DanielleElmas [232]3 years ago
4 0
Give me some answer choices and i will be happy to help
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An object has a mass of 5 kg. What force is needed to accelerate it at 6 m/s2? (Formula: F=ma)
Studentka2010 [4]
Force needed is 5×6=30
5 0
4 years ago
Read 2 more answers
A boy weighs 40 kilograms. He runs at a velocity of 4 meters per second north. Which is his momentum?
Artist 52 [7]
The formula for momentum is mass times velocity. Simply, we just multiply the given values:
p = mv
p = 40 kg x 4 m/s
p = 160 kg m/s

Other units for momentum is N s.
p = 160 N s 
4 0
3 years ago
How do you calculate the net force when there are multiple forces in different directions?​
Artyom0805 [142]

To find F_{net} we need to use vector addition and use the x and y components. First we subtract vector 2 from vector 5 which results in a vector with a  length of 3 pointing directly east, then we use the distance formula to find the length of the net force F_{net} = \sqrt{(3)^2+(4)^2} \\  which gives F_{net} = 5. We now have a magnitude but we also need a direction, since vector 4 and vector 5 are perpendicular. Using \theta = \tan^{-1} (\frac{4}{3})   where tan^-1(y/x) we get an angle of 53 degrees. The resultant force vector is 5 distance with an angle of 53 degrees north east.

4 0
3 years ago
Help ME QUICK PLEASE QUESTION 2 AND 4 PLEASE 58 POINTS
monitta

Answer:

For number 4: A vector pointing to the right with a magnitude of 2.0

Explanation:

Very simple- just subtract 6-2

I am not sure how to do #2- sorry!

6 0
3 years ago
Zad. 2 W marszobiegu chłopiec przebiegł 900 m w ciągu 205 min, a następnie przeszedł 300m w tym danym czasie. Jaka była średnia
USPshnik [31]

Odpowiedź:

0,049 m / s

Wyjaśnienie:

Biorąc pod uwagę, że:

Dystans biegu = 900m

Czas trwania = 205 minut

Długość przejścia = 300 m

Zajęty czas = 205 minut

Średnia prędkość :

(Przebieg + pokonany dystans) / całkowity czas

Średnia prędkość :

(900 m +. 300 m) / 205 + 205

1200 m / 410 minut

Minuty do sekund

1200 / (410 * 60)

1200/24600

= 0,0487804

= 0,049 m / s

8 0
3 years ago
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