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
When hard stabilization structures such as groins are used to stabilize a shoreline, the change in the longshore current results <u>deposition of sediment. </u>
On the upcurrent side of the barrier, sediment is deposited as the longshore current slows.
What is Hard stabilization?
- Hard stabilization is the prevention of erosion through the use of artificial barriers.
- Other hard stabilization structures, such as breakwaters and seawalls, are built parallel to the beach to protect the coast from the force of waves.
- Hard stabilization structures, such as groins, are built at right angles to the shore to prevent the movement of sand down the coast and maintain the beach.
- These constructions are made to last for many years, but because they detract from the visual splendor of the beach, they are not always the ideal answer.
- Additionally, they affect the habitats and breeding sites of native shoreline species, interfering with the ecosystem's natural processes.
Learn more about the Hard stabilization with the help of the given link:
brainly.com/question/16022736
#SPJ4
Answer:
The maximum detection range is 39.75 km
Explanation:
Given that;
Antenna height h1 = 40 m
Target height ( patrol boat mast ) h2 = 11 m
Using RADAR, what is the maximum detection range = ?
Using RADAR
we know that; Maximum detection range = (√17h1 + √17h2) km
where h1 and h2 are heights of the antenna and target height in meters
so we substitute in our values
Maximum detection range = (√(17 × 40) + √(17 × 11)) km
Maximum detection range = (√680 + √187) km
Maximum detection range = (26.0768 + 13.6747) km
Maximum detection range = (26.0768 + 13.6747) km
Maximum detection range = 39.75 km
Therefore, The maximum detection range is 39.75 km
