Divers working deep in the ocean breathe a mixture of gases, of which 97 % is helium [12]. the speed of sound in helium is 965 m
1 answer:
The speed of sound in normal air is approximately 
. In helium, instead, it is 
. The mixture of gases mentioned in the problem is made of 97 % of helium and 3% of normal air, therefore the speed of sound in this mixture will be
So, the ratio between the speed of sound in this mixture of gases and in normal air is
The frequencies of the formants of the diver's voices are proportional to the speed of sound, therefore they will change by the same proportion as the speed of sound does. So, the frequencies will be 2.76 times higher than in normal air.
So, the ratio between the speed of sound in this mixture of gases and in normal air is
The frequencies of the formants of the diver's voices are proportional to the speed of sound, therefore they will change by the same proportion as the speed of sound does. So, the frequencies will be 2.76 times higher than in normal air.
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Answer:
Acceleration,
Explanation:
It is given that,
Initial velocity of the car, u = 10 m/s (in right)
Final velocity of the car, v = -5 m/s (in left)
Time taken, t = 10 s
Let a is the acceleration of the car. It can be calculated using the equation of kinematics. The equation is as :
So, the acceleration of the car is . Hence, this is the required solution.
Answer:
13.6 cm
Explanation:
From Snell's law:
n₁ sin θ₁ = n₂ sin θ₂
In the air, n₁ = 1, and light from the horizon forms a 90° angle with the vertical, so sin θ₁ = sin 90° = 1.
Given n₂ = 4/3:
1 = 4/3 sin θ
sin θ = 3/4
If x is the radius of the circle, then sin θ is:
sin θ = x / √(x² + 12²)
sin θ = x / √(x² + 144)
Substituting:
3/4 = x / √(x² + 144)
9/16 = x² / (x² + 144)
9/16 x² + 81 = x²
81 = 7/16 x²
x ≈ 13.6
