Answer:
critical stress = 595 MPa
Explanation:
given data
fracture toughness = 74.6 MPa-
crack length = 10 mm
f = 1
solution
we know crack length = 10 mm
and crack length = 2a as given in figure attach
so 2a = 10
a = 5 mm
and now we get here with the help of plane strain condition , critical stress is express as
critical stress = 
......................1
put here value and we get
critical stress = 
critical stress = 595 MPa
so here stress is change by plane strain condition because when plate become thinner than condition change by plane strain to plain stress.
plain stress condition occur in thin body where stress through thickness not vary by the thinner section.
Answer:
specific energy = 2.65 ft
y2 = 1.48 ft
Explanation:
given data
average speed v = 6.5 ft/s
width = 5 ft
depth of the water y = 2 ft
solution
we get here specific energy that is express as
specific energy = y + 
...............1
put here value and we get
specific energy = 
specific energy = 2.65 ft
and
alternate depth is
y2 = 
and
here Fr² = 
Fr² = 0.8025
put here value and we get
y2 = 
y2 = 1.48 ft
Answer:
c. 25 μA
Explanation:
The dependent current source means that 31 times i1 will flow through the 6kΩ resistor, effectively increasing its voltage drop to 31 times that which would result from i1 alone. In effect, the 6kΩ resistor behaves in the left-side circuit loop as though it were 31×6kΩ = 186kΩ (with no dependent current source).
Then the current i1 is equivalent to that created by a 5+1 = 6V source through a 54kΩ +186kΩ = 240kΩ circuit impedance.
(6V)/(240kΩ) = 25 μA
_____
<em>Additional comment</em>
The voltage across the 6kΩ resistor is (186/240)·6V = 4.65V, and the 25 μA current generates a voltage of 30·(25 μA)(1.8kΩ) = 1.35V across the 1.8kΩ resistor. This means the voltage source at the right side of the diagram needs to be at least 4.65 +1.35 = 6.0V in order to support the calculated voltage drops.
A good visual lead is 20-30 seconds from the front of the vehicle, focusing in the center of the path of travel. Searching 20 to 30 seconds ahead, gives you time to assess within the next ..12-15 seconds, actions you may need to take to control an approaching risk.
Discussion:
Keeping eye focus centered in a path of travel at an interval of 20 to 30 seconds away from the vehicle is critical to gaining enough info. as possible in the driving scene. Good targeting sets up good sight lines for referencing and good peripheral fields for observing changes.
- It is important to look ahead 12-15 seconds into your target area as one drives. Compromise. space by giving as much space to the greater of two hazards.
Read more on driving visual leads:
brainly.com/question/7067386
