Answer:
Red line is reflected ray.
Explanation:
According to the given figure, the blue line represent the incident ray. It strikes the surface and reflection occurs. The dotted line shows normal.
After reflecting the surface, the ray of light is reflected. Red line shows the reflected light.
In this case, the laws of reflection follows i.e. the angle of incidence is equal to the angle of reflection.
Answer:
a) V ≈ 125 m/s; b) Δt = 13.24 s; c) ΔS ≈ 1450 m
Explanation:
a) We have just to calculate the vector resultant.
V² = 106² + 66.2²
V² = 15618.44
V ≈ 125 m/s
b) The time of flight is equal to the time to reach the maximum height summed to the time to reach the land.
In vertical:
V = V₀ + a * t
V = 66.2 - g * t
0 = 66.2 - 9.8 * t
t ≈ 6.76 s
So: Δt = 13.24 s
c) In horizontal:
V = ΔS / Δt
106 = ΔS / 13.52 ⇒ ΔS = 106 * 13.52
ΔS = 106 * 13.52
ΔS = 1433,12
ΔS ≈ 1450 m
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.