Complete Question
A wave is described by y(x,t) = 0.1 sin(3x + 10t), where x is in meters, y is in centimetres and t is in seconds. The angular wave frequency is
Answer:
The value is ![w = 10 \ rad /s](https://tex.z-dn.net/?f=w%20%3D%20%2010%20%5C%20rad%20%2Fs)
Explanation:
From the question we are told that
The equation describing the wave is y(x,t) = 0.1 sin(3x + 10t)
Generally the sinusoidal equation representing the motion of a wave is mathematically represented as
![y(x,t) = Asin(kx + wt )](https://tex.z-dn.net/?f=y%28x%2Ct%29%20%3D%20%20Asin%28kx%20%2B%20wt%20%29)
Where w is the angular frequency
Now comparing this equation with that given we see that
![w = 10 \ rad /s](https://tex.z-dn.net/?f=w%20%3D%20%2010%20%5C%20rad%20%2Fs)
143m/s if you just perhaps by what you know you'll figure it out
Ok so if each side is 4.53 cm, we can multiply 4.53 x 4.53 x 4.53 to get the volume (since v= l x w x h). Density equals mass/volume, so
519 g/4.53 cm
114.57 g/cm^3 (since none of the units cancel)
1) Frequency: ![3.29\cdot 10^{15}Hz](https://tex.z-dn.net/?f=3.29%5Ccdot%2010%5E%7B15%7DHz)
the energy of the photon absorbed must be equal to the ionization enegy of the atom, which is
![E=2.18 aJ=2.18\cdot 10^{-18} J](https://tex.z-dn.net/?f=E%3D2.18%20aJ%3D2.18%5Ccdot%2010%5E%7B-18%7D%20J)
The energy of a photon is given by
![E=hf](https://tex.z-dn.net/?f=E%3Dhf)
where
is the Planck's constant. By using the energy written above and by re-arranging thsi formula, we can calculate the frequency of the photon:
![f=\frac{E}{h}=\frac{2.18\cdot 10^{-18} J}{6.63\cdot 10^{-34} Js}=3.29\cdot 10^{15} Hz](https://tex.z-dn.net/?f=f%3D%5Cfrac%7BE%7D%7Bh%7D%3D%5Cfrac%7B2.18%5Ccdot%2010%5E%7B-18%7D%20J%7D%7B6.63%5Ccdot%2010%5E%7B-34%7D%20Js%7D%3D3.29%5Ccdot%2010%5E%7B15%7D%20Hz)
2) Wavelength: 91.2 nm
The wavelength of the photon can be found from its frequency, by using the following relationship:
![\lambda=\frac{c}{f}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7Bc%7D%7Bf%7D)
where
is the speed of light and f is the frequency. Substituting the frequency, we find
![\lambda=\frac{3\cdot 10^8 m/s}{3.29\cdot 10^{15}Hz}=9.12\cdot 10^{-8} m=91.2 nm](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7B3%5Ccdot%2010%5E8%20m%2Fs%7D%7B3.29%5Ccdot%2010%5E%7B15%7DHz%7D%3D9.12%5Ccdot%2010%5E%7B-8%7D%20m%3D91.2%20nm)
The intensity on a screen 20 ft from the light will be 0.125-foot candles.
<h3>What is the distance?</h3>
Distance is a numerical representation of the length between two objects or locations.
The intensity I of light varies inversely as the square of the distance D from the source;
I∝(1/D²)
The ratio of the intensity of the two cases;
![\rm \frac{I_1}{I_2} =(\frac{D_2}{D_1} )^2\\\\ \rm \frac{2}{I_2} =(\frac{20}{5} )^2\\\\ \frac{2}{I_2} =4^2 \\\\ I_2= \frac{2}{16} \\\\ I_2= 0.125 \ foot-candles](https://tex.z-dn.net/?f=%5Crm%20%5Cfrac%7BI_1%7D%7BI_2%7D%20%3D%28%5Cfrac%7BD_2%7D%7BD_1%7D%20%29%5E2%5C%5C%5C%5C%20%5Crm%20%5Cfrac%7B2%7D%7BI_2%7D%20%3D%28%5Cfrac%7B20%7D%7B5%7D%20%29%5E2%5C%5C%5C%5C%20%5Cfrac%7B2%7D%7BI_2%7D%20%3D4%5E2%20%5C%5C%5C%5C%20I_2%3D%20%5Cfrac%7B2%7D%7B16%7D%20%5C%5C%5C%5C%20%20I_2%3D%200.125%20%5C%20foot-candles)
Hence, the intensity on a screen 20 ft from the light will be 0.125 foot-candles
To learn more about the distance refer to the link;
brainly.com/question/26711747
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