The wavelength of the infrared radiation is λ = 
×
m.
<h3>What is infrared radiation?</h3>
An infrared telescope is tuned to detect infrared radiation with a frequency of 9.45 THz.
We know that,
1 THz = 10¹² Hz
So,
f = 9.45 × 10¹² Hz
We need to find the wavelength of the infrared radiation.
λ=c/f
λ = 3×
/9.45×
λ = 3.174 × m
The term "infrared radiation" (IR) refers to a part of the electromagnetic radiation spectrum with wavelengths between about 700 nanometers (nm) and one millimeter (mm). Longer than visible light waves but shorter than radio waves are infrared waves.
Electromagnetic radiation with wavelengths longer than those of visible light is known as infrared, also known as infrared light. Since it is undetectable to the human eye, The typical range of wavelengths considered to be infrared (IR) is from about 1 millimeter to the nominal red edge of the visible spectrum, or about 700 nanometers.
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The velocity is given by:
V = √(Vx²+Vy²)
V = velocity, Vx = horizontal velocity, Vy = vertical velocity
Given values:
Vx = 6m/s, Vy = 12m/s
Plug in and solve for V:
V = √(6²+12²)
V = 13.42m/s
Now find the direction:
θ = tan⁻¹(Vy/Vx)
θ = angle of velocity off horizontal, Vy = vertical velocity, Vx = horizontal velocity
Given values:
Vx = 6m/s, Vy = 12m/s
Plug in and solve for θ:
θ = tan⁻¹(12/6)
θ = 63.4°
The resultant velocity is 13.42m/s at an angle of 63.4° off the horizontal.
Momentum (p) = mass × velocity
P= 200,000×4.5
P= 900,000 .... answer !!
Explanation :
It is given that,
BMR i.e basal metabolic rate is 88 kcal/hr. So, BMR in watts is converted by the following :
We know that, 1 kilocalorie = 4184 joules
So, 
J/sec is nothing but watts.
So, 
and 
So, it can be seen that the body can accommodate a modes amount of activity in hot weather but strenuous activity would increase the metabolic rate above the body's ability to remove heat.
The energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b is 3.06 eV.
<h3>Change in energy level of the electron</h3>
When photons jump from a higher energy level to a lower level, they emit or radiate energy.
The change in energy level of the electrons is calculated as follows;
ΔE = Eb - Ef
ΔE = -2.68 eV - (-5.74 eV)
ΔE = 3.06 eV
Thus, the energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b is 3.06 eV.
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