Answer: µ=0.205
Explanation:
The horizontal forces acting on the ladder are the friction(f) at the floor and the normal force (Fw) at the wall. For horizontal equilibrium,
f=Fw
The sum of the moments about the base of the ladder Is 0
ΣM = 0 = Fw*L*sin74.3º - (25.8kg*(L/2) + 67.08kg*0.82L)*cos74.3º*9.8m/s²
Note that it doesn't matter WHAT the length of the ladder is -- it cancels.
Solve this for Fw.
0= 0.9637FwL - (67.91L)2.652
Fw=180.1/0.9637
Fw=186.87N
f=186.81N
Since Fw=f
We know Fw, so we know f.
But f = µ*Fn
where Fn is the normal force at the floor --
Fn = (25.8 + 67.08)kg * 9.8m/s² =
910.22N
so
µ = f / Fn
186.81/910.22
µ= 0.205
Answer:
The tension in the cord is 
Explanation:
Given:
M = mass
b = radius
R = spool of radius
The equation is:
(eq. 1)
The sum of forces in y:
∑Fy = Mg - T = Ma

Replacing in eq. 1

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
Here are the answers to the question. Make sure to give a valid reason, please.
A. the sum of the protons and neutrons in one atom of the element.
B. a ratio based on the mass of a carbon-12 atom.
C. a weighted average of the masses of an element's isotopes.
D. twice the number of protons in one atom of the element.