Answer:
Atoms found in nature are either stable or unstable. ... An atom is unstable (radioactive) if these forces are unbalanced; if the nucleus has an excess of internal energy. Instability of an atom's nucleus may result from an excess of either neutrons or protons
A process known as fixation<span>. the majority of nitrogen is fixed by </span>bacteria<span>, most of which are </span>symbiotic<span> with plants</span>
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Your question has been heard loud and clear.
Well it depends on the magnitude of charges. Generally , when both positive charges have the same magnitude , their equilibrium point is towards the centre joining the two charges. But if magnitude of one positive charge is higher than the other , then the equilibrium point will be towards the charge having lesser magnitude.
Now , a negative charge is placed in between the two positive charges. So , if both positive charges have same magnitude , they both pull the negative charge towards each other with an equal force. Thus the equilibrium point will be where the negative charge is placed because , both forces are equal , and opposite , so they cancel out each other at the point where the negative charge is placed. However if they are of different magnitudes , then the equilibrium point will be shifted towards the positive charge having less magnitude.
Thank you
Answer:
400m
Explanation:
Brainliest? :))
Let your initial displacement from your home to the store be
Dd
>
1 and your displacement from the store to your friend’s house
be Dd
>
2.
Given: Dd
>
1 = 200 m [N]; Dd
>
2 = 600 m [S]
Required: Dd
>
T
Analysis: Dd
>
T 5 Dd
>
1 1 Dd
>
2
Solution: Figure 6 shows the given vectors, with the tip of Dd
>
1
joined to the tail of Dd
>
2. The resultant vector Dd
>
T is drawn in red,
from the tail of Dd
>
1 to the tip of Dd
>
2. The direction of Dd
>
T is [S].
Dd
>
T measures 4 cm in length in Figure 6, so using the scale of
1 cm : 100 m, the actual magnitude of Dd
>
T is 400 m.
Statement: Relative to your starting point at your home, your
total displacement is 400 m [S].