A peak = A Rms x Sq root 2
Therefore 3.6 x sq root of 2
A peak = 5.09
The general formula for the frequency of the nth-harmonic of the column of air in the tube is given by

where f1 is the fundamental frequency.
In our problem, we have two harmonics, one of order n and the other one of order (n+1) (because it is the next higher harmonic), so their frequencies are


so their difference is

So, the difference between the frequencies of the two harmonics is just the fundamental frequency of the column of air in the tube, which is:
I couldn't know for certain cause I don't know what course it is. But according to OSHA, a Exposure Control Plan is used for limited contact with bloody or body fluids so...
A:
is my guess
Answer:
Tբ = (C₁T₁ + C₂T₂)/(C₁ + C₂) = (C₁T₁)/(C₁ + C₂) + (C₂T₂)/(C₁ + C₂)
Explanation:
Let's Assume that T₁ > T₂, this means that, T₁ > Tբ > T₂
The zeroth law of thermodynamics explains that two bodies in thermal equilibrium will eventually end up with the same final body temperature.
Heat lost by body 1 = Heat gained by body 2
C₁ (T₁ - Tբ) = C₂ (Tբ - T₂)
C₁T₁ - C₁Tբ = C₂Tբ - C₂T₂
C₁Tբ + C₂Tբ = C₁T₁ + C₂T₂
Tբ(C₁ + C₂) = C₁T₁ + C₂T₂
Tբ = (C₁T₁ + C₂T₂)/(C₁ + C₂)
Tբ = (C₁T₁)/(C₁ + C₂) + (C₂T₂)/(C₁ + C₂)
Friction will slow down the moving object