Answer:
W = N!/(n0! * n1!)
Explanation:
Let n0 = number of particles in the lowest energy state
n1 = number of particles in the excited energy state.
Using this, we can say that N = n0 + n1
From this we can then express the weight, W of the close system by finding the factorials of each particles
W = N!/(n0! * n1!)
Hence, the weight W is expressed as W = N!/(n0! * n1!)
1. Frequency: 
The frequency of a light wave is given by:

where
is the speed of light
is the wavelength of the wave
In this problem, we have light with wavelength

Substituting into the equation, we find the frequency:

2. Period: 
The period of a wave is equal to the reciprocal of the frequency:

The frequency of this light wave is
(found in the previous exercise), so the period is:

Answer:
Fn: magnitude of the net force.
Fn=30.11N , oriented 75.3 ° clockwise from the -x axis
Explanation:
Components on the x-y axes of the 17 N force(F₁)
F₁x=17*cos48°= 11.38N
F₁y=17*sin48° = 12.63 N
Components on the x-y axes of the the second force(F₂)
F₂x= −19.0 N
F₂y= 16.5 N
Components on the x-y axes of the net force (Fn)
Fnx= F₁x +F₂x= 11.38N−19.0 N= -7.62 N
Fny= F₁y +F₂y= 12.63 N +16.5 N = 29.13 N
Magnitude of the net force.



Direction of the net force (β)

β=75.3°
Magnitude and direction of the net force
Fn= 30.11N , oriented 75.3 ° clockwise from the -x axis
In the attached graph we can observe the magnitude and direction of the net force
The Electromagnetic spectrum.
M)³ / 6 = 4.2e9 m³
<span>so its mass is </span>
<span>M = 3300kg/m³ * 4.2e9m³ = 1.4e13 kg </span>
<span>and so its KE at 16 km/s = 16000 m/s is </span>
<span>KE = ½ * 1.4e13kg * (16000m/s)² = 1.8e21 J
</span># of bombs N = 1.8e21J / 4.0e16J/bomb = 44 234 bombs
<span>give or take.
</span>
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