What is the weighted average mass of chlorine
1 answer:
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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
Answer:
Magnitude of displacement = 2.07 km
Magnitude of average velocity = 1.17 kmph
Explanation:
Let east represent positive x axis and north represent positive y axis.
A bird watcher meanders through the woods, walking 1.93 km due east, 1.03 km due south, and 3.84 km in a direction 52.8 ° north of west.
1.93 km due wast
s ₁ = 1.93 i km
1.03 km due south
s₂ = -1.03 j km
3.84 km in a direction 52.8 ° north of west
s₃ = -3.84 cos 52.8 i + 3.84 sin 52.8 j = -2.32 i + 3.06 j km
Total displacement
s = s ₁+ s₂+ s₃ = 1.93 i - 1.03 j -2.32 i + 3.06 j = -0.39 i + 2.03 j
Magnitude of displacement,
Time taken = 1.771 hour
Magnitude of average velocity,
Answer:
Explanation:
Given
Force of repulsion between two charge particle is given by force F
Electrostatic force is given by
where and
is the charges of particle
r=distance between charge particle
when charges are doubled and distance is reduced to half
i.e. q become 2 q and r becomes 0.5 r