This question is about Circle Geometry. it evaluates connected and broken lines with respect to circles.
<h3>What is Circle Geometry?</h3>
This refers to the body of knowledge in mathematics that has to do with the various problems associated with the Circle.
In real-world scenarios, circle geometry is used in technologies involving:
- Camera lenses
- Circular Architectural structures
- Steering Wheels
- Buttons etc.
Learn more about Circle Geometry at:
brainly.com/question/24375372
Answer:
The answer is below
Explanation:
Given that:
Diameter (D) = 0.03 mm = 0.00003 m, length (L) = 2.4 mm = 0.0024 m, longitudinal tensile strength 
, Fracture strength
a) The critical length (
) is given by:
The critical length (4.5 mm) is greater than the given length, hence th composite can be produced.
b) The volume fraction (Vf) is gotten from the formula:
Answer:
The pressure drop across the pipe also reduces by half of its initial value if the viscosity of the fluid reduces by half of its original value.
Explanation:
For a fully developed laminar flow in a circular pipe, the flowrate (volumetric) is given by the Hagen-Poiseulle's equation.
Q = π(ΔPR⁴/8μL)
where Q = volumetric flowrate
ΔP = Pressure drop across the pipe
μ = fluid viscosity
L = pipe length
If all the other parameters are kept constant, the pressure drop across the circular pipe is directly proportional to the viscosity of the fluid flowing in the pipe
ΔP = μ(8QL/πR⁴)
ΔP = Kμ
K = (8QL/πR⁴) = constant (for this question)
ΔP = Kμ
K = (ΔP/μ)
So, if the viscosity is halved, the new viscosity (μ₁) will be half of the original viscosity (μ).
μ₁ = (μ/2)
The new pressure drop (ΔP₁) is then
ΔP₁ = Kμ₁ = K(μ/2)
Recall,
K = (ΔP/μ)
ΔP₁ = K(μ/2) = (ΔP/μ) × (μ/2) = (ΔP/2)
Hence, the pressure drop across the pipe also reduces by half of its initial value if the viscosity of the fluid reduces by half of its value.
Hope this Helps!!!
Answer:
Yes, fracture will occur
Explanation:
Half length of internal crack will be 4mm/2=2mm=0.002m
To find the dimensionless parameter, we use critical stress crack propagation equation
and making Y the subject
Where Y is the dimensionless parameter, a is half length of crack, K is plane strain fracture toughness, 
is critical stress required for initiating crack propagation. Substituting the figures given in question we obtain
When the maximum internal crack length is 6mm, half the length of internal crack is 6mm/2=3mm=0.003m
and making K the subject
and substituting 260 MPa for 
while a is taken as 0.003m and Y is already known
Therefore, fracture toughness at critical stress when maximum internal crack is 6mm is 42.455 Mpa and since it’s greater than 40 Mpa, fracture occurs to the material
