Answer:
The sound intensity that the other student measure, I₂ is expected to be;
A) 6.2 × 10⁻⁸ W/m²
Explanation:
The measure of sound intensity is given by the following formula;
Where;
I = The intensity
R = The radius
P = The power of the sound
Whereby we have;
The distance of the two people talking, R₁ = 3.0 m
The measure of the sound intensity, I₁ = 1.1 × 10⁻⁷ W/m² (from an online source)
The distance of the other student from the two people talking, R₂ = 4.0 m
Therefore, the estimate of the sound intensity, I₂, is given as follows;
I₂ = 6.1875 × 10⁻⁸ W/m²
∴ The sound intensity that the other student measure, I₂ ≈ 6.2 × 10⁻⁸ W/m²
The cathode and anode are hooked up to an electrical circuit The chemical reactions in the battery causes a build up of electrons at the anode. This results in an electrical difference between the anode and the cathode. Hope this helps :)
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Answer:
λ = 3 10⁻⁷ m, UV laser
Explanation:
The diffraction phenomenon is described by the expression
a sin θ = m λ
let's use trigonometry
tan θ = y / L
as in this phenomenon the angles are small
tan θ = 
= sin θ
sin θ = y / L
we substitute
a y / L = m λ
let's apply this equation to the initial data
a 0.04 / L = 1 600 10⁻⁹
a / L = 1.5 10⁻⁵
now they tell us that we change the laser and we have y = 0.04 m for m = 2
a 0.04 / L = 2 λ
a / L = 50 λ
we solve the two expression is
1.5 10⁻⁵ = 50 λ
λ = 1.5 10⁻⁵ / 50
λ = 3 10⁻⁷ m
UV laser
