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Darina [25.2K]
3 years ago
14

How many electrons does nitrogen (N) need to gain to have a stable outer electron shell?

Physics
1 answer:
Gemiola [76]3 years ago
3 0
It needs 3 electrons to have it stabled
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Can u find the . in this?
nikklg [1K]

Answer:

Yeah it's right there from the one next to the exclamation point

4 0
3 years ago
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The equation for photosynthesis is 6H2O (water) + 6CO2 (carbon dioxide) + Light Energy → C6H12O6 (glucose) + 6O2 (oxygen). When
aleksklad [387]

Answer:

b

Explanation:

bc

8 0
3 years ago
A group of bike riders took a 4 hour trip. During the first 3 hours, they traveled a total of 50 km, but
zlopas [31]

Answer:

15km/hr

Explanation:

The average speed for the entire trip is the sum of the total distance traveled divided by the total time of the trip.

Total time  = 4hr

distance for the first 3hrs  = 50km

distance for the last 1hr  = 10km

  Total distance  = 50km + 10km  = 60km

Now;

  Average speed  = \frac{total distance }{time taken}  

 Insert the parameters and solve;

 Average speed  = \frac{60km}{4hr}   = 15km/hr

7 0
3 years ago
Suppose that the moment of inertia of a skater with arms out and one leg extended is 3.5 kg⋅m2 and for arms and legs in is 0.70
Aleksandr-060686 [28]
Given:
I₁ = 0.70 kg-m², the moment of inertia with arms and legs in
I₂ = 3.5 kg-m², the moment of inertia with arms and a leg out.
ω₁ = 4.8 rev/s, the angular speed with arms and legs in.
That is,
ω₁ = (4.8 rev/s)*(2π rad/rev) = 30.159 rad/s

Let ω₂ =  the angular speed with arms and a leg out.
Because momentum is conserved, therefore
I₂ω₂ = I₁ω₁
ω₂ = (I₁/I₂)ω₁
      = (0.7/3.5)*(30.159)
      = 6.032 rad/s

ω₂ = (6.032 rad/s)*(1/(2π) rev/rad) = 0.96 rev/s

Answer: 0.96 rev/s


3 0
3 years ago
Consider a concave mirror that has a focal length f. In terms of f, determine the object distances that will produce a magnifica
Aleks04 [339]

We have that the magnification of each focal length is given respectively as

A) has u=3\frac{f}{2}

B) has u=4\frac{f}{3}

C) has  u=5\frac{f}{4}

From the question we are told that:

Focal Length F

Generally, the equation for Magnification is mathematically given by

M=\frac{-v}{u}

Therefore

v=2u

For A

M=-2

Therefore

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

\frac{1}{f}=\frac{1}{u}+\frac{1}{2u}

Therefore

u=3\frac{f}{2}

For B

M=-3

Therefore

v=3u

Where

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

\frac{1}{f}=\frac{1}{u}+\frac{1}{3u}

Therefore

u=4\frac{f}{3}

For C

M=-4

Therefore

v=4u

Therefore

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

\frac{1}{f}=\frac{1}{u}+\frac{1}{4u}

Therefore

u=5\frac{f}{4}

Conclusion

From the calculations above we can rightly say that the magnifications of the values above are

A has u=3\frac{f}{2}

B has u=4\frac{f}{3}

C has  u=5\frac{f}{4}

For more information on this visit

brainly.com/question/14468351

3 0
2 years ago
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