Answer:
The height of the water column = 1.62405
× 10⁻¹ m
Explanation:
The air cavity in the Coke bottle = 0.220 m deep
The fundamental (frequency) it plays when water is added to shorten the column and it is blown across the top, f = 528 Hz
The given speed of sound in air, v = 343 m/s
We note that the air cavity in the coke bottle is equivalent to a tube closed at one end
The fundamental frequency for a tube closed at one end, 'f', is given as follows;
f = v/(4·L) = v/λ
Where;
L = The height of the water column
λ = The wavelength of the wave
∴ 4·L = v/f = (343 m/s)/(528 Hz) = 0.6496
m
∴ L = 0.6496 m/4 = 0.162405 m
The height of the water column = 1.62405 × 10⁻¹ m.
Answer:
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Explanation:
Answer:
Explanation:
Assuming the squirrel is jumping off the ground, here's what we know but don't really know...
v₀ = 4.0 at 50.0°
So that's not really the velocity we are looking for. We are dealing with a max height problem, which is a y-dimension thing. Therefore, we need the squirrel's upward velocity, which is NOT 4.0 m/s. We find it in the following way:
which gives us that the upward velocity is
v₀ = 3.1 m/s
Moving on here's what we also know:
a = -9.8 m/s/s and
v = 0
Remember that at the very top of the parabolic path, the final velocity is 0. In order to find the max height of the squirrel, we need to know how long it took him to get there. We are using 2 of our 3 one-dimensional equations in this problem. To find time:
v = v₀ + at and filling in:
0 = 3.1 - 9.8t and
-3.1 = -9.8t so
t = .32 seconds.
Now that we know how long it took him to get to the max height, we use that in our next one-dimensional equation:
Δx = 
and filling in:
Δx = 
and using the rules for adding and subtracting sig fig's correctly, we can begin to simplify this:
Δx = .99 - .50 so
Δx = .49 meters
Answer:
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Answer:
When a light wave goes through a slit, it is diffracted, which means the slit opening acts as a new source of waves. How much a light wave diffracts<em> (how much it fans out)</em> depends on the wavelength of the incident light. The wavelength must be larger than the width of the slit for the maximum diffraction. Thus, for a given slit, red light, because it has a longer wavelength, diffracts more than the blue light.
The corresponding relation for diffraction is
,
where 
is the wavelength of light, 
is the slit width, and 
is the diffraction angle.
From this relation we clearly see that the diffraction angle is directly proportional to the wavelength of light—longer the wavelength larger the diffraction angle.
