Answer:
Each of the metal specimens HAS an indentation near the center to ensure that the fracture point would occur in this region. Tension tests WERE CONDUCTED as follows. Two pieces of reflective tape WERE PLACED approximately 1 inch apart in the center of the specimen where the indentation 4 WAS LOCATED. The width and the thickness of the specimen at this location WAS MEASURED using a Vernier caliper. Then the specimen WAS SECURED in the MTS Load Frame. A laser extensometer WAS PLACED into position to measure the deformation of the specimen. The laser extensometer WAS USED to measure the original distance between the pieces of reflective tape. The MTS WAS SET to elongate the specimen one tenth of an inch every minute.
Answer:
Its data
Explanation:
just took the assignment on edge 2021
Answer:
676 ft
Explanation:
Minimum sight distance, d_min
d_min = 1.47 * v_max * t_total where v_max is maximum velocity in mi/h, t_total is total time
v_max is given as 50 mi/h
t_total is sum of time for right-turn and adjustment time=8.5+0.7=9.2 seconds
Substituting these figures we obtain d_min=1.47*50*9.2=676.2 ft
For practical purposes, this distance is taken as 676 ft
Answer:288 pm
Explanation:
Number of atoms(s) for face centered unit cell -
Lattice points: at corners and face centers of unit cell.
For face centered cubic (FCC), z=4.
- whereas
For an FCC lattices √2a =4r =2d
Therefore d = a/√2a = 408pm/√2a= 288pm
I think with this step by step procedure the, the answer was clearly stated.
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