Answer:
It will not experience fracture when it is exposed to a stress of 1030 MPa.
Explanation:
Given
Klc = 54.8 MPa √m
a = 0.5 mm = 0.5*10⁻³m
Y = 1.0
This problem asks us to determine whether or not the 4340 steel alloy specimen will fracture when exposed to a stress of 1030 MPa, given the values of <em>KIc</em>, <em>Y</em>, and the largest value of <em>a</em> in the material. This requires that we solve for <em>σc</em> from the following equation:
<em>σc = KIc / (Y*√(π*a))</em>
Thus
σc = 54.8 MPa √m / (1.0*√(π*0.5*10⁻³m))
⇒ σc = 1382.67 MPa > 1030 MPa
Therefore, the fracture will not occur because this specimen can handle a stress of 1382.67 MPa before experience fracture.
Answer: D
Explanation: When you change your transmission fluid you change your filter. When you do that you clean where the filter was. Then you can put the new filter on and the new fluid.
Answer:
B. Hybrid
Explanation:
A hybrid desing implicate combining two different elements/methods, such a hybrid car, which works using both mechanic and electric energy.
Answer:
Δ enthalpy = -23 Btu/Ibm
Explanation:
Given data:
Pressure ( P1 ) = 250 psi
Initial Temperature ( T1 ) = 175°F
Final temperature ( T2 ) = 20°F
<u>Calculate the change in the enthalpy of R-134a </u>
From R-134 table
h1 = 129.85 Btu/Ibm
s1 = 0.23281 Btu/Ibm.R
note : entropy is constant
hence ; s1 = s2
by interpolation ; h2 = 106.95
Δ enthalpy = h2 - h1
= ( 106.95 - 129.85 ) = -23 Btu/Ibm
Answer:
engine
Explanation:
as long as the engine and evrything is running it should be good
