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
Answer:
The change in momentum is
Explanation:
From the question we are told that
The mass of the probe is 
The location of the prob at time t = 22.9 s is 
The momentum at time t = 22.9 s is
The net force on the probe is 
Generally the change in momentum is mathematically represented as

The initial time is 22.6 s
The final time is 22.9 s
Substituting values

Answer:
<em>113.4 J</em>
Explanation:
<u>Elastic Potential Energy</u>
Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

The spring has a natural length of 0.7 m and a spring constant of k=70 N/m. When the spring is stretched to a length of 2.5 m, the change of length is
Δx = 2.5 m - 0.7 m = 1.8 m
The energy stored in the spring is:

PE = 113.4 J
11m if you add 6+5 you get 11 but of course you need the “m” in the mix so 11m but correct me if I’m wrong.