Answer:
at load 13114.85 N beam fail
Explanation:
given data
radius r = 6.7 mm
load = 2710 N
distance = 41 mm
length = 19 mm
to find out
At what load we expect this specimen to fracture
solution
we will first flexural strength that is
flexural strength = load × distance / πr³
flexural strength = 2710 × 41 ×
/ π(6.7 ×)³
flexural strength = 117.592 MPa
we know that for cross section specimen
flexural strength = 3load × distance / 2bd²
put here these value
117.592 × 
N/m² = 3 load × 41 × / ( 2 × (19×)³)
load = 117.592 × × ( 2 × (19×)³) / ( 3 × 41 × )
load = 13114.85 N
so at load 13114.85 N beam fail
Answer:
A disk 8.00 cm in radiu: rotates at a constant rate of 1200 revinin about its central axis Determine (a) its angular speed in zadians per second, (b) the tangential speed at a point 3.00 cm from its center, (c) the radial aceleration of a point on the rim, and (d) the total distance a point our tke rim noves ign-2.00 s (E) The moment of inertia if it's mass is 2Kg? is the answer
Answer:
70,100 m/s
Explanation:
Given:
v₀ = 205 m/s
a = 8.03 m/s²
Δx = 1750 m
Find: v
v² = v₀² + 2aΔx
v² = (205 m/s)² + 2 (8.03 m/s²) (1750 m)
v = 70,130 m/s
Rounding to three significant figures, the final velocity is 70,100 m/s.
