Answer:
88750 N
Explanation:
given data:
plastic deformation σy=266 MPa=266*10^6 N/m^2
cross-sectional area Ao=333 mm^2=333*10^-6 m^2
solution:
To determine the maximum load that can be applied without
plastic deformation (Fy).
Fy=σy*Ao
=88750 N
Answer:
L= 50000 lb
D = 5000 lb
Explanation:
To maintain a level flight the lift must equal the weight in magnitude.
We know the weight is of 50000 lb, so the lift must be the same.
L = W = 50000 lb
The L/D ratio is 10 so
10 = L/D
D = L/10
D = 50000/10 = 5000 lb
To maintain steady speed the thrust must equal the drag, so
T = D = 5000 lb
Answer:
critical stress required for the propagation is 27.396615 ×
N/m²
Explanation:
given data
specific surface energy = 0.90 J/m²
modulus of elasticity E = 393 GPa = 393 ×
N/m²
internal crack length = 0.6 mm
to find out
critical stress required for the propagation
solution
we will apply here critical stress formula for propagation of internal crack
( σc ) =
.....................1
here E is modulus of elasticity and γs is specific surface energy and a is half length of crack i.e 0.3 mm = 0.3 ×
m
so now put value in equation 1 we get
( σc ) =
( σc ) =
( σc ) = 27.396615 ×
N/m²
so critical stress required for the propagation is 27.396615 ×
N/m²
Answer:
The answer is "
".
Explanation:
Please find the correct question in the attachment file.
using formula:



Answer:
It creates airflow in the engine
Explanation: