Your friend has two substances A and B which are compressed liquid and superheated vapor respectively. Both are in rigid vessels
1 answer:
Answer:
Following are the solution to the given question:
Explanation:
The material is necessary to cook since frying is a speedy process for evaporation.
Drug A is now in the compressed fluid area, and the material would not boil if the pressure is chilled. Because the ship is solid, its substance A claim is false.
Unless the volume comprises of a drop in the heat, the B substance reaches a vapor pressure area and a wet region. That's the area that melting may occur. His claim for material B could therefore be true.
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Answer:
Explanation:
In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.
We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.
The weigth depends on the size and specific gravity.
W = 1/2 * b * h * L * SG
Then
Teq = 1/2 * b * h * L * SG * b / 2
Teq = 1/4 * b^2 * h * L * SG
The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:
The term sin(30) is because of the slope of the wall
The pressure of water is:
p(y) = SGw * (h - y)
Then:
T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)
T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)
T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)
T1 = 1/3 * SGw * sin(30) * L * h^3
To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)
1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3
In an equilateral triangle h = b * cos(30)
1/4 * b^3 * cos(30) * L * SG > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3
SG > SGw * 4/3* sin(30) * (cos(30))^2
SG > 1/2 * SGw
For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.
This is avergae specific gravity, including holes.
Answer:
Step 1 of 3
Case A:
AISI 1018 CD steel,
Fillet radius at wall=0.1 in,
Diameter of bar
From table deterministic ASTM minimum tensile and yield strengths for some hot rolled and cold drawn steels for 1018 CD steel
Tensile strength
Yield strength
The cross section at A experiences maximum bending moment at wall and constant torsion throughout the length. Due to reasonably high length to diameter ratio transverse shear will be very small compared to bending and torsion.
At the critical stress elements on the top and bottom surfaces transverse shear is zero
Explanation:
See the next steps in the attached image
