Answer:
Remember that the nucleus of an athom is neutral, which means protons, neutrons and electrons have the same and opposite charge so that the net electircal field is zero. In that case, we say the nucleus is estable.
Now, to mantain the equilibrium of the given element, we need an electrical field equal in magintude +2e with opposite direction, that will balance it.
Answer:
v = -v₀ / 2
Explanation:
For this exercise let's use kinematics relations.
Let's use the initial conditions to find the acceleration of the electron
v² = v₀² - 2a y
when the initial velocity is vo it reaches just the negative plate so v = 0
a = v₀² / 2y
now they tell us that the initial velocity is half
v’² = v₀’² - 2 a y’
v₀ ’= v₀ / 2
at the point where turn v = 0
0 = v₀² /4 - 2 a y '
v₀² /4 = 2 (v₀² / 2y) y’
y = 4 y'
y ’= y / 4
We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.
v² = v₀² -2a y’
v² = 0 - 2 (v₀² / 2y) y / 4
v² = -v₀² / 4
v = -v₀ / 2
We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.
Answer:
1) Q ’= 8 Q
, 2) q ’= 16 q
, 3) r ’= ¾ r
Explanation:
For this exercise we will use Coulomb's law
F = k q Q / r²
It asks us to calculate the change of any of the parameters so that the force is always F
Original values
q, Q, r
Scenario 1
q ’= 2q
r ’= 4r
F = k q ’Q’ / r’²
we substitute
F = k 2q Q ’/ (4r)²
F = k 2q Q '/ 16r²
we substitute the value of F
k q Q / r² = k q Q '/ 8r²
Q ’= 8 Q
Scenario 2
Q ’= Q
r ’= 4r
we substitute
F = k q ’Q / 16r²
k q Q / r² = k q’ Q / 16 r²
q ’= 16 q
Scenario 3
q ’= 3/2 q
Q ’= ⅜ Q
we substitute
k q Q r² = k (3/2 q) (⅜ Q) / r’²
r’² = 9/16 r²
r ’= ¾ r
All of them will have the same density
Answer: vf= 364 m/s
Explanation: Use the formula for acceleration:
a= vf - vi / t
Derive for vf:
vf =( at )+ vi
= 5.4 m/s² ( 55s) + 67 m/s
= 364 m/s
