Answer:
B. In mercury, the frequency of the wave is the same as in ethanol, but the wavelength is greater.
Explanation:
To solve this easily, we can just calculate the wavelength of the sound in Ethanol and in Mercury.
In Ethanol, the wavelength will be:
λ = c/f
λ = 1160/440
λ = 2.63 m
In Mercury, the wavelength will be:
λ = c/f
λ = 1450/440
λ = 3.3 m
The wavelength of sound is greater in Mercury than in Ethanol but the frequency is the same.
Frequency of sound is not dependent on medium, but velocity and wavelength change depending on the medium.
Answer:
Frequency.
Explanation:
Sound are mechanical waves that are highly dependent on matter for their propagation and transmission.
Sound travels faster through solids than it does through either liquids or gases. A student could verify this statement by measuring the time required for sound to travel a set distance through a solid, a liquid, and a gas.
Mathematically, the speed of a sound is given by the formula:
Speed = wavelength * frequency
The pitch of a sound you hear depends on the frequency of the sound wave.
Hmmm, no choices? No? Okay then... I believe it should be 9.0 J.
Answer:
Explanation:
Answer: Let ke = 1/2 IW^2 = 1/2 kMr^2 W^2 be Earth's rotational KE. W = 2pi/24 radians per hour rotation speed and k = 2/5 for a solid sphere M is Earth mass, r = 6.4E6 m.
Then ke = 1/2 2/5 6E24 (6.4E6)^2 (2pi/(24*3600))^2 = ? Joules. You can do the math, note W is converted to radians per second for unit consistency.
Let KE = 1/2 KMR^2 w^2 be Earth's orbital KE. w = 2pi/(365*24) radians per hour K = 1 for a point mass. Note I used 365 days, a more precise number is 365.25 days per year, which is why we have Leap Years.
Find KE/ke = 1/2 KMR^2 w^2//1/2 kMr^2 W^2 = (K/k)(w/W)^2 (R/r)^2 = (5/2) (365)^2 (1.5E11/6.4E6)^2 = 7.81E9 ANS
The answer to your question is A.