Answer:
2.08V/cm
Explanation:
Plot the points on a graph. Draw the line of best fit. Calculate the gradietn of line of best fit.
Attached is the graph plotted on excel.
The equation of the line is
Votlage= 2.08× distance + 0.276
The shaft is hollow from A to B and solid from B to C. The shaft has an outer diameter of 79 mm, and the thickness of the wall o
f the hallow segment is 10 mm.
Determine the maximum shear stress developed in the shaft. Express your answer to three significant figures and include appropriate units.
Determine the maximum shear stress developed in the shaft. Express your answer to three significant figures and include appropriate units.
1 answer:
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Answer:
The answer is below
Explanation:
Let A represent the first switch, B represent the second switch and C represent the bulb. Also, let 0 mean turned off and 1 mean turned on. Since when both switches are in the same position, the light is off. This can be represented by the following truth table:
A B C (output)
0 0 0
0 1 1
1 0 1
1 1 0
The logic circuit can be represented by:
C = A'B + AB'
The output (bulb) is on if the switches are at different positions; if the switches are at the same position, the output (bulb) is off. This is an XOR gate. The gate is represented in the diagram attached below.
Answer:
correct option is (A) 0.5
Explanation:
given data
axial column load = 250 kN per meter
footing placed = 0.5 m
cohesion = 25 kPa
internal friction angle = 5°
solution
we know angle of internal friction is 5° that is near to 0°
so it means the soil is almost cohesive soil.
and for a pure cohesive soil
= 0
and we know formula for is
= (Nq - 1 ) × tan(Ф) ..................1
so here Ф is very less should be nearest to zero
and its value can be 0.5
so correct option is (A) 0.5
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