The optimum wavelength is 450 nm because that is the wavelength of maximum absorbance by FeSCN2+(aq)
you should choose a wavelength with maximum absorbance. In this case, you are using the scattered light, not the absorbed light as your signal. So you should avoid wavelengths where there are absorption peaks.
<h3>What is wavelength ?</h3>
A waveform signal that is carried in space or down a wire has a wavelength, which is the separation between two identical places (adjacent crests) in the consecutive cycles. This length is typically defined in wireless systems in metres (m), centimetres (cm), or millimetres (mm) (mm).
- The distance between two waves' crests serves as an illustration of wavelength. When you and another person have the same overall mindset and can easily communicate, that is an example of being on the same wavelength.
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Answer:
The force of static friction acting on the luggage is, Fₓ = 180.32 N
Explanation:
Given data,
The mass of the luggage, m = 23 kg
You pulled the luggage with a force of, F = 77 N
The coefficient of static friction of luggage and floor, μₓ = 0.8
The formula for static frictional force is,
Fₓ = μₓ · η
Where,
η - normal force acting on the luggage 'mg'
Substituting the values in the above equation,
Fₓ = 0.8 x 23 x 9.8
= 180.32 N
Hence, the minimum force require to pull the luggage is, Fₓ = 180.32 N
Answer:
25 m/s
Explanation:
First we should define the variables
T=4
Dx = 100
ay=-9.8
ax=0
We can use formula 1 from the BIG 5
x=(v+v0)t/2
By plugging in our variables we can get 100=4(v+v0)/2
Which is 50=v+v0
v=v0 since horizontal acceleration always equals zero
so 2v0 = 50
v0 = 25
(a) The work done by the applied force is 26.65 J.
(b) The work done by the normal force exerted by the table is 0.
(c) The work done by the force of gravity is 0.
(d) The work done by the net force on the block is 26.65 J.
<h3>
Work done by the applied force</h3>
W = Fdcosθ
W = 14 x 2.1 x cos25
W = 26.65 J
<h3>
Work done by the normal force</h3>
W = Fₙd
W = mg cosθ x d
W = (2.5 x 9.8) x cos(90) x 2.1
W = 0 J
<h3>Work done force of gravity</h3>
The work done by force of gravity is also zero, since the weight is at 90⁰ to the displacement.
<h3> Work done by the net force on the block</h3>
∑W = 0 + 26.65 J = 26.65 J
Thus, the work done by the applied force is 26.65 J.
The work done by the normal force exerted by the table is 0.
The work done by the force of gravity is 0.
The work done by the net force on the block is 26.65 J.
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