<span>The speed of a wave, V, is f *lambda. Where f is the frequency and lambda is the distance. If a new crest reaches the end every 4 secs; it takes 8s to cover the distance. Hence, f, which is the number of oscillations covered is 8s. So we have V = 8 * 5 = 40 ms^1.</span>
Answer:
4miles/hour
Explanation:
the solution for this question requires that the quantities are converted to the appropriate units as required by the question.
Rate in miles per hour = distance in miles / time in hour
to convert 12 minutes to hours; recall that 60 minutes make 1 hour
12 minutes to hour = 12/60 = 0.2hr
to convert 4224 feet to miles; recall 5280 feet is equivalent to 1 mile
4224 feet to miles = 4224/5280 = 0.8 miles
∴ rate = 0.8 / 0.2
rate = 4 miles per hour
the constant rate in miles per hour = 4 miles/hour
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).
Hello There!
The most accurate statement is B.
Hope This Helps You!
Good Luck :)