We will first determine using the given if an aircraft component will fracture with a given stress level (260 MPa), maximum internal crack length (6.0 mm) and fracture toughness (40 MPa m ), given that fracture occurs for the same component using the same alloy for another stress level and internal crack length. First, it is necessary to solve for the parameter Y, using Equation 8.5, for the conditions under which fracture occurred (i.e., σ = 300 MPa and 2 a = 4.0 mm). Therefore,
Y = K(Ic)/ sqrt(π a) = 40 MPa( m ) / (300 MPa) sqrt(( π ) ((4 × 10-3 m)/2)) = 1.68
We will now solve for the product Y σ π a for the other set of conditions, so as to ascertain whether or not this value is greater than the K(Ic) for the alloy. Thus,
Y sqrt(π a) = (1.68)(260 MPa) sqrt (( π )[(6 × 10^-3 m)/ 2])
= 42.4 MPa sqrt (m) (39 ksi in. )
Therefore, fracture will occur since this value ( 42.4 MPa sqrt(m)) is greater than the K(Ic) of the material, 40 MPa sqrt(m).
<span>Nodes are points along the medium that appear to be standing still. They are points on a standing wave that has no displacement from the rest position.
Antinodes are the opposite of nodes. Those are</span> points that undergo the maximum displacement.
<span>At the nodes the destructive interference occurs</span> . For example <span>a crest of one wave meets a trough of a second wave</span> , or a half-crest of one wave meets a half-trough of a second wave...
Answer:
the angular acceleration of the car is 1.5 rad/s²
Explanation:
Given;
initial angular velocity, 
= 10 rad/s
final angular velocity, 
= 25 rad/s
time of motion, t = 10 s
The angular acceleration of the car is calculated as follows;
Therefore, the angular acceleration of the car is 1.5 rad/s²
B; to produce a hypothesis.
