Answer:
ΔS=2*m*Cp*ln((T1+T2)/(2*(T1*T2)^1/2))
Explanation:
The concepts and formulas that I will use to solve this exercise are the integration and the change in the entropy of the universe. To calculate the final temperature of the water the expression for the equilibrium temperature will be used. Similarly, to find the change in entropy from cold to hot water, the equation of the change of entropy will be used. In the attached image is detailed the step by step of the resolution.
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
A. Use the thin lens equation to determine the image distance from the lens. Is the correct answer because I have no idea why
We know, the ideal gas equation,
P1V1 / T1 = P2V2 / T2
Here, P1 = 760 mm
V1 = 10 m3
T1 = 27 + 273 = 300 K
P2 = 400 mm Hg
T2 = -23 + 273 = 250 K
Substitute their values,
760*10 / 300 = 400 * V2 / 250
25.33 * 250 = 400 * V2
V2 = 6333.333/ 400
V2 = 15.83
In short, Your Answer would be approx. 15.83 m3
Hope this helps!