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:
Only Technician B is right.
Explanation:
The cylindrical braking system for a car works through the mode of pressure transmission, that is, the pressure applied to the brake pedals, is transmitted to the brake pad through the cylindrical piston.
Pressure applied on the pedal, P(pedal) = P(pad)
And the Pressure is the applied force/area for either pad or pedal. That is, P(pad) = Force(pad)/A(pad) & P(pedal) = F(pedal)/A(pedal)
If the area of piston increases, A(pad) increases and the P(pad) drops, Meaning, the pressure transmitted to the pad reduces. And for most cars, there's a pressure limit for the braking system to work.
If the A(pad) increases, P(pad) decreases and the braking force applied has to increase, to counter balance the dropping pressure and raise it.
This whole setup does not depend on the length of the braking lines; it only depends on the applied force and cross sectional Area (size) of the piston.
Answer:
Im guessing this is for CEA for PLTW, if so look up the exact assignment number and look at online examples of the exact same assignment.
Explanation:
