Answer:
λ = 396.7 nm
Explanation:
For this exercise we use the diffraction ratio of a grating
d sin θ = m λ
in general the networks works in the first order m = 1
we can use trigonometry, remembering that in diffraction experiments the angles are small
tan θ = y / L
tan θ = 
= sin θ
sin θ = y / L
we substitute
= m λ
with the initial data we look for the distance between the lines
d = 
d = 1 656 10⁻⁹ 1.00 / 0.600
d = 1.09 10⁻⁶ m
for the unknown lamp we look for the wavelength
λ = d y / L m
λ = 1.09 10⁻⁶ 0.364 / 1.00 1
λ = 3.9676 10⁻⁷ m
λ = 3.967 10⁻⁷ m
we reduce nm
λ = 396.7 nm
Errrrr, it has 80.
80 is the correct answer
Distance of lake a is 200 km at 20 degree north of east
distance between lake a and b is 230 km at 30 degree west of north
now the distance between base and lake b is given as
given that
now the total distance is
now the magnitude of the distance is given as
also the direction is given as
<em>so it is 277.4 km at 74.7 degree North of East</em>
Answer:
The answer to your question is:
a) t = 3.81 s
b) vf = 37.4 m/s
Explanation:
Data
height = 71.3 m = 234 feet
t = 0 m/s
vf = ?
vo = 0 m/s
Formula
h = vot + 1/2gt²
vf = vo + gt
Process
a)
h = vot + 1/2gt²
71.3 = 0t + 1/2(9.81)t²
2(71.3) = 9,81t²
t² = 2(71.3)/9.81
t² = 14.53
t = 3.81 s
b)
vf = 0 + (9.81)(3.81)
vf = 37.4 m/s
Answer:
P = 7196 [kPa]
Explanation:
We can solve this problem using the expression that defines the pressure depending on the height of water column.
P = dens*g*h
where:
dens = 1028 [kg/m^3]
g = 10 [m/s^2]
h = 700 [m]
Therefore:
P = 1028*10*700
P = 7196000 [Pa]
P = 7196 [kPa]
