Answer:
λ = 548.7 nm
Explanation:
Hi!
First we want to know how much energy we need to remove 1 electron from the surface of the solid:
218.1 kJ/mol => 218 100 J / (6.022 x 10^23) electrons
= 3.621 x 10^-19 J/electron
That is we need 3.621 x 10^-19 J to remove one electron
Now we can calculate the wavelength that a photon must have in order to have this energy:
E = (hc) / λ
λ = (hc) / Ε
where
h = 6,626070150(69) ×10 -34 Js (wikipedia)
c = 3 x10^8 m/s
hc = 1.987 x 10^-25 Jm
Therefore:
λ = ( 1.987 x 10^-25 /3.621 x 10^-19 ) m = 5.487 x 10^-7 m
λ = 548.7 nm
Answer:
sorry
Explanation:
I dont know the anwser to that one
Answer:
Explanation:
The force of kinetic friction on the block is defined as:
Where 
is the coefficient of kinetic friction between the block and the surface and N is the normal force, which is always perpendicular to the surface that the object contacts. So, according to the free body diagram of the block, we have:
Replacing this in the first equation and solving for :
Answer:
55.5 degrees above the x-axis.
Explanation:
Using SOH-CAH-TOA we can determine that
tan(the angle) = (opposite side)/(adjacent side)
Inserting our variables we get
the angle = arctan(28.4/19.5)
the angle = 55.5 degrees
Answer:
a) F = 35.7 N, b) W = 846.7 J, c) W = - 846.9 J, d) W=0
Explanation:
a) For this exercise let's use Newton's second law, let's set a reference frame with the x-axis horizontally
let's break down the pushing force.
cos (-23,7) = Fₓ / F
sin (-237) = F_y / F
Fₓ = F cos 23.7 = F 0.916
F_y = F sin (-23.7) = - F 0.402
Y axis
N- W - F_y = 0
N = W + F 0.402
X axis
Fₓ - fr = 0
F 0.916 = fr
F = fr / 0.916
F = 32.7 / 0.916
F = 35.7 N
It is asked to calculate several jobs
b) the work of the pushing force
W = fx x
W = 35.7 cos 23.7 25.9
W = 846.7 J
c) friction force work
W = F x cos tea
friction force opposes movement
W = - fr x
W = - 32.7 25.9
W = - 846.9 J
d) The work of the force would gravitate, as the displacement and the force of gravity are at 90º, the work is zero
W = 0
