Answer:
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Explanation:
Answer:
The magnitude of the electric field between the membranes is 1.13 x 10⁶ N/C
Explanation:
Given;
the distance of separation of the parallel plate capacitor, d = 10nm
the charge density, σ = 10⁻⁵C/m²
the magnitude of the electric field between the membranes, is calculated using the formula below;
E = σ / ε₀
Where;
ε₀ is permittivity of free space, = 8.85 x 10⁻¹² C²/Nm²
E is magnitude of the electric field between the membranes
σ is surface charge density
E = (10⁻⁵) / (8.85 x 10⁻¹²)
E = 1.13 x 10⁶ N/C
Therefore, the magnitude of the electric field between the membranes is 1.13 x 10⁶ N/C
Answer:
Explanation:
All objects <u>radiate</u> energy.
I believe that the answer is true!
Answer
given,
time = 10 s
ship's speed = 5 Km/h
F = m a
a is the acceleration and m is mass.
In the first case
F₁=m x a₁
where a₁ = difference in velocity / time
F₁ is constant acceleration is also a constant.
Δv₁ = 5 x 0.278
Δv₁ = 1.39 m/s

a₁ = 0.139 m/s²
F₂ =m x a₂
F₃ = F₂ + F₁
Δv₃ = 19 x 0.278
Δv₃ = 5.282 m/s
a₃=Δv₂ / t

a₃ = 0.5282 m²/s
m a₃=m a₁ + m a₂
a₃ = a₂ + a₁
0.5282 = a₂ + 0.139
a₂=0.3892 m²/s
F₂ = m x 0.3892...........(1)
F₁ = m x 0.139...............(2)
F₂/F₁
ratio = 
ratio = 2.8