Being able to see a square in the middle of the image despite not having lines to form a square represents the Gestalt principle
2 answers:
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Answer:
The particle in the D ring is 1399 times faster than the particle in the Encke Division.
Explanation:
The circular velocity is define as:
Where r is the radius of the trajectory and T is the orbital period
To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:
Where T is orbital period and r is the radius of the trajectory.
Case for the particle in the Encke Division:
It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( )
1 AU is defined as the distance between the earth and the sun.
But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.
That can be expressed in units of days
<em>Circular velocity for the particle in the </em><em>Encke Division</em><em>:</em>
For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.
⇒ 10247040 s
⇒ 133370000 m
Case for the particle in the D Ring:
For the case of the particle in the D Ring, the same approach used above can be followed
<em>Circular velocity for the particle in </em><em>D Ring</em><em>:</em>
For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.
⇒ 3786912 s
⇒ 69000000 m
The particle in the D ring is 1399 times faster than the particle in the Encke Division.
Answer:
a) ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J, d) W = 150 10⁻¹² J
Explanation:
Let's find the capacitance of the capacitor
C =
C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³
C = 2.62 10⁻¹² F
for the initial data let's look for the accumulated charge on the plates
C =
Q₀ = C ΔV
Q₀ = 2.62 10⁻¹² 8.70
Q₀ = 22.8 10⁻¹² C
a) we look for the capacity for the new distance
C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³
C₁ = 1.04 10⁻¹² F
C₁ = Q₀ / ΔV₁
ΔV₁ = Q₀ / C₁
ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²
ΔV₁ = 21.9 V
b) initial stored energy
U₀ =
U₀ = (22.8 10⁻¹²)²/(2 2.62 10⁻¹²)
U₀ = 99.2 10⁻¹² J
c) final stored energy
U_f = (22.8 10⁻¹²) ² /(2 1.04 10⁻⁻¹²)
U_f = 249.9 10⁻¹² J
d) the work of separating the plates
as energy is conserved work must be equal to energy change
W = U_f - U₀
W = (249.2 - 99.2) 10⁻¹²
W = 150 10⁻¹² J
note that as the energy increases the work must be supplied to the system