Answer:
#_photon = 5 10²⁰ photons / s
Explanation:
For this exercise let's calculate the energy of a single quantum of energy, use Planck's law
E = h f
c= λ f
E = h c / λ
λ= 1000 nm (1 m / 109 nm) = 1000 10⁻⁹ m
Let's calculate
E₀ = 6.6310⁻³⁴ 3 10⁸/1000 10⁻⁹
E₀ = 19.89 10⁻²⁰ J
This is the energy emitted by a photon let's use a proportions rule to find the number emitted in P = 100 w
#_photon = P / E₀
#_photon = 100 / 19.89 10⁻²⁰
#_photon = 5 10²⁰ photons / s
I would have to see the graph.. but by looking at one one online, they are between points D and E.
Answer:
The velocity of the ball before it hits the ground is 381.2 m/s
Explanation:
Given;
time taken to reach the ground, t = 38.9 s
The height of fall is given by;
h = ¹/₂gt²
h = ¹/₂(9.8)(38.9)²
h = 7414.73 m
The velocity of the ball before it hits the ground is given as;
v² = u² + 2gh
where;
u is the initial velocity of the on the root = 0
v is the final velocity of the ball before it hits the ground
v² = 2gh
v = √2gh
v = √(2 x 9.8 x 7414.73 )
v = 381.2 m/s
Therefore, the velocity of the ball before it hits the ground is 381.2 m/s
The Kelvin scale has no negatives on it.
Zero Kelvin is 'Absolute Zero', and nothing can get colder than that.
