Problem 3. The uniform beam is supported by two rods AB and CD that have cross-sectional areas of 10 mm2 and 15mm2, respectively
1 answer:
Answer:
hello the diagram related to your question is missing attached below is the required diagram.
The value of AC = 6m
answer 2.25 N
Explanation:
From the diagram attached below
average intensity of loading = w/2
Total load due to linearly varying load =
w = maximum intensity of load , L = span of varying load
since AC = 6m( L ) then
Total load due to linearly varying load = 3w ------ ( 1 )
attached below is the detailed solution of the given problem
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Answer:
kinematic viscosity is 0.0149 ft²/s
Explanation:
given data
specific gravity S = 0.94
density ρ = 0.94 × 1000
viscosity μ = 13 Poise = 1.3 Pa-sec
we know 1 poise = 0.1 pas
to find out
kinematic viscosity
solution
we will apply here Kinematic viscosity formula that is
kinematic viscosity = ..................1
put here value in equation 1
and here ρ is density and μ is viscosity
kinematic viscosity =
kinematic viscosity = 1.382978 × m³/s
so kinematic viscosity is 0.0149 ft²/s
Complete Question
The complete question is shown on the first uploaded image
Answer:
The solution is shown on the second, third ,fourth
Explanation:
The explanation is shown on the second, third image
Answer:
Explanation:
See the attached picture for detailed answer.
