The temperature of the cosmic background radiation is measured to be 2.7 K. What is the wavelength of the peak in the spectral distribution at this temperature?
What frequency corresponds to this wavelength?
A 112 g hockey puck glides across a frictionless ice surface with no horizontal forces acting on it. If the puck's velocity is 22.5 m/s to the right at t = 0 s, what will the puck's horizontal velocity be at t = 225 ms? Round your answer to the nearest 0.1 m/s.
25 m/s to the right
To calculate the wavelength of the peak in the spectral distribution, we can use Wien's displacement law. According to Wien's law, the wavelength peak (λpeak) is inversely proportional to the temperature (T). The formula is given as follows:
λpeak = (2.898 * 10^-3 m·K) / T
To find the wavelength of the peak at a temperature of 2.7 K, we can substitute this value into the equation:
λpeak = (2.898 * 10^-3 m·K) / 2.7 K
= (2.898 * 10^-3) / 2.7
≈ 1.073 * 10^-3 meters
The wavelength of the peak in the spectral distribution at a temperature of 2.7 K is approximately 1.073 * 10^-3 meters.
To determine the frequency corresponding to this wavelength, we can use the equation:
c = λ * ν
where c represents the speed of light, λ is the wavelength, and ν denotes the frequency.
Rearranging the equation, we have:
ν = c / λ
The speed of light (c) is approximately 3.00 * 10^8 meters per second. Substituting the known values, we get:
ν = (3.00 * 10^8 m/s) / (1.073 * 10^-3 m)
≈ 2.796 * 10^11 Hz
Therefore, the frequency corresponding to a wavelength of approximately 1.073 * 10^-3 meters is approximately 2.796 * 10^11 Hz.