It’s c “ the same size as you are”
<h2>Answer: It becomes an Ion
</h2>
When an atom has gained or lost electrons (negative charge), it becomes an ion.
In this sense:
<h2>I
ons are atoms that have <u>
gained or lost</u>
electrons in their electronic cortex.
</h2><h2>
</h2>
If a neutral atom <u>loses electrons</u>, it remains with an excess of positive charge and transforms into a positive ion or <u>cation</u>, whereas if a neutral atom <u>gains electrons</u>, it acquires an excess of negative charge and transforms into a negative ion or <u>anion</u>.
It is then how ions form bonds with other atoms differently depending on the number of electrons they have.
Answer:
a) -2.516 × 10⁻⁴ V
b) -1.33 × 10⁻³ V
Explanation:
The electric field inside the sphere can be expressed as:
The potential at a distance can be represented as:
V(r) - V(0) = 
V(r) - V(0) = ![[\frac{qr^2}{8 \pi E_0R^3 }]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bqr%5E2%7D%7B8%20%5Cpi%20E_0R%5E3%20%7D%5D)
₀
V(r) = ![-[\frac{qr^2}{8 \pi E_0R^3 }]](https://tex.z-dn.net/?f=-%5B%5Cfrac%7Bqr%5E2%7D%7B8%20%5Cpi%20E_0R%5E3%20%7D%5D)
₀
Given that:
q = +3.83 fc = 3.83 × 10⁻¹⁵ C
r = 0.56 cm
= 0.56 × 10⁻² m
R = 1.29 cm
= 1.29 × 10⁻² m
E₀ = 8.85 × 10⁻¹² F/m
Substituting our values; we have:
= -2.15 × 10⁻⁴ V
The difference between the radial distance and center can be expressed as:
V(r) - V(0) = 
V(r) - V(0) = ![[\frac{qr^2}{8 \pi E_0R^3 }]^R](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bqr%5E2%7D%7B8%20%5Cpi%20E_0R%5E3%20%7D%5D%5ER)
V(r) = 
V(r) = 
V(r) 
V(r) = -0.00133
V(r) = - 1.33 × 10⁻³ V
Answer:
I don't understand please interpret
Answer:
.
Explanation:
By Newton's Second Law, the acceleration 
of an object is proportional to the net force 
on it. In particular, if the mass of the object is 
, then
.
Rewrite this equation to obtain:
.
In this case, the assumption is that the 
force is the only force that is acting on the object. Hence, the net force on the object would also be
Make sure that all values are in their standard units. Forces should be in Newtons (same as 
, and the acceleration of the object should be in meters-per-second-squared (
). Apply the equation to find the mass of the object.
.
