Folds in the bone they help warm and moisten incoming air.
1 answer:
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Answer:
Speed is same as that before it entered glass.
Explanation:
Given:
A light ray enters and passes through the glass as shown in the diagram.
We have to analyze its speed.
Speed of light in air is and speed of light in glass is
Whenever a light ray enters a glass block or slab there is bending of light at the interface of the two media.
So speed of light will decrease in glass medium and again it passes to the air.
So
Speed of light in air will again increase or will be equivalent to the earlier speed when it was entering the glass block.
Finally
Speed is same as that before it entered glass as it in the same medium (air).
Complete question:
A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).
Required:
What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.
Answer:
The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m
Explanation:
Given;
charge of the coaxial capable, Q = 8.5 µC = 8.5 x 10⁻⁶ C
length of the conductor, L = 50 m
inner radius, r₁ = 1.304 mm
outer radius, r₂ = 9.249 mm
The magnitude of the electric field halfway between the two cylindrical conductors is given by;
Where;
λ is linear charge density or charge per unit length
r is the distance halfway between the two cylindrical conductors
The magnitude of the electric field is now given as;
Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m
Answer:
a) I = 3.63 W / m² , b) I = 0.750 W / m²
Explanation:
The intensity of a sound wave is given by the relation
I = P / A = ½ ρ v (2π f )²
I = (½ ρ v 4π² s_{max}²) f²
a) with the initial condition let's call the intensity Io
cte = (½ ρ v 4π² s_{max}²)
I₀ = cte s² f₀²
I₀ = cte 10 6
If frequency is increase f = 2.20 10³ Hz
I = constant (2.20 10³) 2
I = cte 4.84 10⁶
let's find the relationship of the two quantities
I / Io = 4.84
I = 4.84 Io
I = 4.84 0.750
I = 3.63 W / m²
b) in this case the frequency is reduced to f = 0.250 10³ Hz and the displacement s = 4 s or
I = cte (f s)²
I = constant (0.250 10³ 4)²
I = cte 1 10⁶
the relationship
I / Io = 1
I = Io
I = 0.750 W / m²