You gently place several steel needles on the free surface of the water in a large tank. The needles come in two lengths: some a
re 5 cm long, and some are 10 cm long. Needles of each length are available with diameters of 1 mm, 2.5 mm, and 5 mm.
Make a prediction as to which needles, if any will float.
Assume contact angle theta = 0. (For water, rho=1000kg/m3, sigma=0.0728 N/m, for steel, SG= 0.83).
Make a prediction as to which needles, if any will float.
Assume contact angle theta = 0. (For water, rho=1000kg/m3, sigma=0.0728 N/m, for steel, SG= 0.83).
1 answer:
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Explanation:
Styrene is a vinyl monomer in which there is a carbon carbon double bond.
The polymerization of the styrene, which is initiated by using a free radical which reacts with the styrene and the compound thus forms react again and again to form polystyrene (PS).
The equation is shown below as:
⇒
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