Answer:
1) The speed of sound increases
2) 440 Hz
3) 29°C
4) 17°C
5) 434 Hz
6) 12 m/s
7) 17.3 m
Explanation:
1) The speed of sound increases
2) V = f×λ
f = V/λ = 343/0.78 = 439.744 ≈ 440 Hz
3) V = f×λ
512 × 0.68 = 348.16 m/s
348.16 - 331 = 17.16
T = 17.16/0.6 = 28.6 ≈ 29°C
4) Increase in speed = 350 - 340 = 10
Increase in temperature = 10/0.6 = 16.67° ≈ 17°C
5) f = V/λ = 343/0.79 = 434 Hz
6) 331 + 0.6×30 - (331 × 0.6 ×10) = 12 m/s
7) V = 331 + 0.6×25 = 346m/s
λ = 346/20 = 17.3 m
Answer:
1.
2.
3.The results from part 1 and 2 agree when r = R.
Explanation:
The volume charge density is given as
We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.
1. Since the cylinder is very long, Gauss’ Law can be applied.
The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is
where ‘h’ is the length of the imaginary Gaussian surface.
2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,
3. At the boundary where r = R:
As can be seen from above, two E-field values are equal as predicted.