Answer:
So, at the depth of 24 cm below the surface of the glycerine the pressure is 2970 Pa. Hence, this is the required solution.
Explanation:
Given that,
Pressure exerted by the surface of glycerine, P = 2970 Pa and it is greater than atmospheric pressure.
The density of glycerine,
We need to find the depth h below the surface of the glycerine. The pressure due to some depth is given by :
h = 0.24 meters
or
h = 24 cm
At STP, 1 mole of an ideal gas occupies a volume of about 22.4 L. So if <em>n</em> is the number of moles of this gas, then
<em>n</em> / (19.2 L) = (1 mole) / (22.4 L) ==> <em>n</em> = (19.2 L•mole) / (22.4 L) ≈ 0.857 mol
If the sample has a mass of 12.0 g, then its molecular weight is
(12.0 g) / <em>n</em> ≈ 14.0 g/mol
Answer:
The first minimum would be observed at 41.57°
Explanation:
v = 340m/s = speed of sound
f = 610Hz
d = 0.840m
λ = ?
Mλ = wsinθ
m = mth order minima
λ = wavelength incident on the single slit
θ = angular position of the mth minima
But, λ = v / f
λ = 340 / 610 = 0.557m
θ = sin⁻(mλ/d)
θ = sin⁻ [(1 * 0.557) / 0.840]
θ = sin⁻ 0.6635
θ = 41.57°
The first minimum would be observed at 41.57°
Answer: 0.85 meters (with and without sigfigs)
Explanation: To find the wavelength, you just have to switch around the equation for wave speed: v (wave speed) = λ (wavelength)*f (frequency) so λ (wavelength) = v (wave speed)/f (frequency). You don't have the wave speed but you can calculate it. Since wave speed is measured in meters/second or m/s, you just have to divide the amount of meters you were given by the amount of seconds. You will get 340 m/s. Next, you have to plug the values into the equation: λ (wavelength) = 340 m/s (wave speed)/400 Hz (frequency). The answer is 0.85 meters (seconds cancel) and has the correct number of significant figures.
Omitting the 1 will not change the value of the number, but will change the units at the end of the problem