The range of frequencies of visible light in a vacuum is mathematically given as
Fmin=4.19*10^14Hz to Fmax=1*10^15Hz
<h3>What is the range of frequencies of visible light in a vacuum?</h3>
Question Parameters:
The wavelengths of visible light vary from about 300 nm to 700 nm.
Generally, the equation for the frequency is mathematically given as
F=C/\lambda
Therefore
For Fmax

Fmax=1*10^15Hz
Where

Fmin=4.19*10^14Hz
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Answer:
Kinetic energy is the energy due to motion. Potential energy is energy stored in matter. The joule (J) is the SI unit of energy and equals (kg×m2s2) ( kg × m 2 s 2 ) .
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Answer:
129 J/Kg°C
Explanation:
Given :
Mass of gold, m = 1.2kg
Quantity of heat applied, Q = 3096 J
Temperature, t2 = 40°C
Temperature, t1 = 20°C
Change in temperature, dt = (40-20)°C = 20°C
Using the relation :
Q = mcdt
Where, C = specific heat capacity of gold
3096 = 1.2kg * C * 20°C
3096 J = 24kg°C * C
C = 3096 J / 24 kg°C
C = 129 J/Kg°C