So as you may know atoms are neutral because the number of protons (+ charge) and the number of electrons( - charge) are the same so they cancel out. When a valence electron leaves an atom it will have a +1 charge because there is one less negative than positives or there is one more positive than negatives since a negative electron left. If a valence electron is added a -1 charge because there is now one more negative than positive!!!
hope that helps!!
Answer:
A) K / K₀ = 4 b) v / v₀ = 4
Explanation:
A) For this exercise we can use the conservation of mechanical energy
in the problem it indicates that the displacement was doubled (x = 2xo)
starting point. At the position of maximum displacement
Em₀ = Ke = ½ k (2x₀)²
final point. In the equilibrium position
= K = ½ m v²
Em₀ = Em_{f}
½ k 4 x₀² = K
(½ K x₀²) = K₀
K = 4 K₀
K / K₀ = 4
B) the speed value
½ k 4 x₀² = ½ m v²
v = 4 (k / m) x₀
if we call
v₀ = k / m x₀
v = 4 v₀
v / v₀ = 4
The directions of the vectors for velocity and acceleration are in the opposite directions.
- The velocity vector is always in the direction of motion of the object. So, the direction of velocity is in the right from our point of view.
- When there is a positive acceleration in the object the acceleration vector is in the direction of motion of the object. When there is a negative acceleration in the object the acceleration vector is in the opposite direction of motion of the object. So, the direction of velocity is in the left from our point of view.
Velocity vector is the rate of change of position of an object. Acceleration vector is the rate of change of velocity of an object.
Therefore, the directions of the vectors for velocity and acceleration are in the opposite directions.
To know more about velocity and acceleration vectors
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Answer:
p = 1.16 10⁻¹⁴ C m and ΔU = 2.7 10 -11 J
Explanation:
The dipole moment of a dipole is the product of charges by distance
p = 2 a q
With 2a the distance between the charges and the magnitude of the charges
p = 1.7 10⁻⁹ 6.8 10⁻⁶
p = 1.16 10⁻¹⁴ C m
The potential energie dipole is described by the expression
U = - p E cos θ
Where θ is the angle between the dipole and the electric field, the zero value of the potential energy is located for when the dipole is perpendicular to the electric field line
Orientation parallel to the field
θ = 0º
U = 1.16 10⁻¹⁴ 1160 cos 0
U1 = 1.35 10⁻¹¹ J
Antiparallel orientation
θ = 180º
cos 180 = -1
U2 = -1.35 10⁻¹¹ J
The difference in energy between these two configurations is the subtraction of the energies
ΔU = | U1 -U2 |
ΔU = 1.35 10-11 - (-1.35 10-11)
ΔU = 2.7 10 -11 J
