Acceleration=velocity/time
acceleration=28/4.22
therefore, acceleration=6.64
Answer:
f3 = 102 Hz
Explanation:
To find the frequency of the sound produced by the pipe you use the following formula:
![f_n=\frac{nv_s}{4L}](https://tex.z-dn.net/?f=f_n%3D%5Cfrac%7Bnv_s%7D%7B4L%7D)
n: number of the harmonic = 3
vs: speed of sound = 340 m/s
L: length of the pipe = 2.5 m
You replace the values of n, L and vs in order to calculate the frequency:
![f_{3}=\frac{(3)(340m/s)}{4(2.5m)}=102\ Hz](https://tex.z-dn.net/?f=f_%7B3%7D%3D%5Cfrac%7B%283%29%28340m%2Fs%29%7D%7B4%282.5m%29%7D%3D102%5C%20Hz)
hence, the frequency of the third overtone is 102 Hz
A) We want to find the work function of the potassium. Apply this equation:
E = 1243/λ - Φ
E = energy of photoelectron, λ = incoming light wavelength, Φ = potassium work function
Given values:
E = 2.93eV, λ = 240nm
Plug in and solve for Φ:
2.93 = 1243/240 - Φ
Φ = 2.25eV
B) We want to find the threshold wavelength, i.e. find the wavelength such that the energy E of the photoelectrons is 0eV. Plug in E = 0eV and Φ = 2.25eV and solve for the threshold wavelength λ:
E = 1243/λ - Φ
0 = 1243/λ - Φ
0 = 1243/λ - 2.25
λ = 552nm
C) We want to find the frequency associated with the threshold wavelength. Apply this equation:
c = fλ
c = speed of light in a vacuum, f = frequency, λ = wavelength
Given values:
c = 3×10⁸m/s, λ = 5.52×10⁻⁷m
Plug in and solve for f:
3×10⁸ = f(5.52×10⁻⁷)
f = 5.43×10¹⁴Hz
Explanation:
Submultiples Multiples
Value SI symbol Name
10−1 s ds decasecond
10−2 s cs hectosecond
10−3 s ms kilosecond
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