Answer:
λ = 5.65m
Explanation:
The Path Difference Condition is given as:
δ=
;
where lamda is represent by the symbol (λ) and is the wavelength we are meant to calculate.
m = no of openings which is 2
∴δ= 
n is the index of refraction of the medium in which the wave is traveling
To find δ we have;
δ= 
δ= 
δ= 
δ= 
δ= 
δ= 
δ= 82.15 -73.68
δ= 8.47
Again remember; to calculate the wavelength of the ocean waves; we have:
δ=
δ= 8.47
8.47 =
λ = 
λ = 5.65m
The pressure exerted by the concrete cylinder is 2.60 pound/in².
We need to know about the pressure to solve this problem. Pressure is a unit that describes how much force is applied to a surface area. It can be determined as
P = F / A
where P is pressure, F is force and A is area.
From the question above, we know that
F = 375 pound
A = 144 in²
By substituting the given parameters, we can calculate the pressure
P = F / A
P = 375 / 144
P = 2.60 pound/in²
Thus, the pressure should be 2.60 pound/in².
Find more on pressure at: brainly.com/question/25965960
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Yes it would get colder because people help people learn through experiences with no one to help teach the world would be a cold lifeless place.
Ok so here is the thing. It is necessary to introduce the atomic number Z into the following equation and the reason for that is that we are not working here with hydrogen (H). It will go like this:
<span>E=(2.18×10^-18 J)(Z^2 )|1/(ni^2 )-1/(nf^2 )| </span>
<span>E=(2.18×10^-18 J)(2^2 )|1/(6 ^2 )-1/(4 ^2 )|=3.02798×10^-19 J </span>
<span>After that we need to plug the E value calculated into the equation. Remember that the wavelength is always positive:</span>
<span>E=hc/λ 3.02798×10^-19 J=hc/λ λ=6.56×10^-7 m </span>
so 6.56×10^-7 m or better written 656 nm is in the visible spectrum
Answer:
1. True WA > WB > WC
Explanation:
In this exercise they give work for several different configurations and ask that we show the relationship between them, the best way to do this is to calculate each work separately.
A) Work is the product of force by distance and the cosine of the angle between them
WA = W h cos 0
WA = mg h
B) On a ramp without rubbing
Sin30 = h / L
L = h / sin 30
WB = F d cos θ
WB = F L cos 30
WB = mf (h / sin30) cos 30
WB = mg h ctan 30
C) Ramp with rubbing
W sin 30 - fr = ma
N- Wcos30 = 0
W sin 30 - μ W cos 30 = ma
F = W (sin30 - μ cos30)
WC = mg (sin30 - μ cos30) h / sin30
Wc = mg (1 - μ ctan30) h
When we review the affirmation it is the work where there is rubbing is the smallest and the work where it comes in free fall at the maximum
Let's review the claims
1. True The work of gravity is the greatest and the work where there is friction is the least
2 False. The job where there is friction is the least
3 False work with rubbing is the least
4 False work with rubbing is the least
