Answer:
88750 N
Explanation:
given data:
plastic deformation σy=266 MPa=266*10^6 N/m^2
cross-sectional area Ao=333 mm^2=333*10^-6 m^2
solution:
To determine the maximum load that can be applied without
plastic deformation (Fy).
Fy=σy*Ao
=88750 N
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Answer:
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Explanation:
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Answer:
The average thickness of the blubber is<u> 0.077 m</u>
Explanation:
Here, we want to calculate the average thickness of the Walrus blubber.
We employ a mathematical formula to calculate this;
The rate of heat transfer(H) through the Walrus blubber = dQ/dT = KA(T2-T1)/L
Where dQ is the change in amount of heat transferred
dT is the temperature gradient(change in temperature) i.e T2-T1
dQ/dT = 220 W
K is the conductivity of fatty tissue without blood = 0.20 (J/s · m · °C)
A is the surface area which is 2.23 m^2
T2 = 37.0 °C
T1 = -1.0 °C
L is ?
We can rewrite the equation in terms of L as follows;
L × dQ/dT = KA(T2-T1)
L = KA(T2-T1) ÷ dQ/dT
Imputing the values listed above;
L = (0.2 * 2.23)(37-(-1))/220
L = (0.2 * 2.23 * 38)/220 = 16.948/220 = 0.077 m