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
Answer:
Q = 4.52 10¹⁷ J
Explanation:
Thermal energy can be calculated with
Q = m c_{e} ΔT
in this case it indicates that we approximate seawater to pure water with
c_{e} = 4186 J/ kg K
with the density
ρ = m / V
m = ρ V
V = L³
we substitute
m = ρ L³
Q = ρ L3 c_{e} ΔT
calculate
Q = 1000 (3 103) 3 4186 4
Q = 4.52 10¹⁷ J
Explanation:
There are five equations of motion:
v = at + v₀
Δx = v₀ t + ½ at²
Δx = ½ (v + v₀)t
v² = v₀² + 2aΔx
Δx = vt − ½ at²
Δx is the displacement
v₀ is the initial velocity
v is the final velocity
a is the acceleration
t is time
Answer:
is there any choice? in this question?