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
Answer:
Kindly check the explanation section.
Explanation:
For the design we are asked for in this question/problem there is the need for us to calculate or determine the strength in fracture and that of the yield. Also, we need to calculate for the block shear strength.
From the question, we have that the factored load = 500kips. Also, note that the tension splice must not slip.
Also, the shear force are resisted by friction, that is to say shear resistance = 1.13 × Tb × Ns.
Assuming our db = 3/4 inches, then the slip critical resistance to shear service load = 18ksi(refer to AISC manual for the table).
If db = 7/8 inches, then the shear force resistance for n bolt = 10.2kips, n > 49.6.
The yielding strength = 0.9 × Aj × Fhb= 736 kips > 500
The fracture strength = .75 × Ah × Fhb = 309 kips.
The bearing strength of 7/8 inches bolt at the edge hole and other holes = 46 kips and 102 kips.
Answer:
distance must be = 330 × 5/2
= 330×2.5
=825m
Answer:
D. Validity group
Explanation:
A validity group is not necessarily a component of a normal experiment.
A normal experiment is made up of the dependent variables, independent variables, and most times the control group.
- The dependent variable is one that relies on the outcome of the effect produced from the experimental design
- The independent variable is the one that causes a desired effect under the study.
- The control group is a group that does not receive any treatment nor attenuation of the conditions.