Answer:
C.
Explanation:
If u really need it, just comment.
Answer:
It should be b hope it helps :)
Explanation:
Answer:
a. d₁/d₂ = 1.09 b. 0.054 mW
Explanation:
a. What is the ratio of the diameter of the first student's eardrum to that of the second student?
We know since the power is the same for both students, intensity I ∝ I/A where A = surface area of ear drum. If we assume it to be circular, A = πd²/4 where r = radius. So, A ∝ d²
So, I ∝ I/d²
I₁/I₂ = d₂²/d₁² where I₁ = intensity at eardrum of first student, d₁ = diameter of first student's eardrum, I₂ = intensity at eardrum of second student, d₂ = diameter of second student's eardrum.
Given that I₂ = 1.18I₁
I₂/I₁ = 1.18
Since I₁/I₂ = d₂²/d₁²
√(I₁/I₂) = d₂/d₁
d₁/d₂ = √(I₂/I₁)
d₁/d₂ = √1.18
d₁/d₂ = 1.09
So, the ratio of the diameter of the first student's eardrum to that of the second student is 1.09
b. If the diameter of the second student's eardrum is 1.01 cm. how much acoustic power, in microwatts, is striking each of his (and the other student's) eardrums?
We know intensity, I = P/A where P = acoustic power and A = area = πd²/4
Now, P = IA
= I₂A₂
= I₂πd₂²/4
= 1.18I₁πd₂²/4
Given that I₁ = 0.58 W/m² and d₂ = 1.01 cm = 1.01 × 10⁻² m
So, P = 1.18I₁πd₂²/4
= 1.18 × 0.58 W/m² × π × (1.01 × 10⁻² m)²/4
= 0.691244π × 10⁻⁴ W/4 =
2.172 × 10⁻⁴ W/4
= 0.543 × 10⁻⁴ W
= 0.0543 × 10⁻³ W
= 0.0543 mW
≅ 0.054 mW
Answer:
Explanation:
Let assume that gases inside bulbs behave as an ideal gas and have the same temperature. Then, conditions of gases before and after valve opened are now modelled:
Bulb A (2 L, 2 atm) - Before opening:
Bulb B (3 L, 4 atm) - Before opening:
Bulbs A & B (5 L) - After opening:
After some algebraic manipulation, a formula for final pressure is derived:
And final pressure is obtained:
