Answer:
elongation of the brass rod is 0.01956 mm
Explanation:
given data
length = 5 cm = 50 mm
diameter = 4.50 mm
Young's modulus = 98.0 GPa
load = 610 N
to find out
what will be the elongation of the brass rod in mm
solution
we know here change in length formula that is express as
δ = 
................1
here δ is change in length and P is applied load and A id cross section area and E is Young's modulus and L is length
so all value in equation 1
δ =
δ = 
δ = 0.01956 mm
so elongation of the brass rod is 0.01956 mm
Answer:per minute from the pumping well, a steady state was attained in about 24 hr. The draw-down at a distance of 10 ft. was 5.5 ft. and at 25 ft. was 1.21 ft.
Explanation:
Answer:
a) Cr = Co - Fx / D
b) ΔC / Δx = ( CR - Cr ) / ( xR - xRo )
Explanation:
A) Derive an expression for the profile c(r) inside the tissue
F = DΔC / X = D ( Co - Cr ) / X ------ 1
where : F = flux , D = drug diffusion coefficient
X = radial distance between Ro and R
Hence : Cr = Co - Fx / D
B) Express the diffusive flux at outer surface of the balloon
Diffusive flux at outer surface = ΔC / Δx = CR - Cr / xR - xRo
Answer:
peak flow and any engineering considerations related thereto
Explanation:
It should be no surprise that a peak flow meter will report peak flow, sometimes with important maximum-value, time-constant, or bandwidth limitations. There are many engineering issues related to flow rates. A peak flow meter can allow you to assess those issues with respect to the flows actually encountered.
Peak flow can allow you to assess adequacy of flow and whether there may be blockages or impediments to flow that reduce peak levels below expected values. An appropriate peak flow meter can help you assess the length of time that peak flow can be maintained, and whether that delivers sufficient volume.
It can also allow you to assess whether appropriate accommodation is made for unexpectedly high flow rates. (Are buffers or overflow tanks of sufficient size? Is there adequate protection against possible erosion? Is there adequate support where flow changes direction?)
The smallest area of each cable if the stress is not to exceed 90MPa in bronze is 43.6 mm² and 120MPa in steel is 32.7 mm².
<h3>What is normal stress?</h3>
If the direction of deformation force is perpendicular to the cross-sectional area of the body, the stress is called normal stress. Changes in wire length and body volume will be normal.
σ = P/A
Where, σ = Normal stress
P = Pressure
A = Area
1 Kg = 9.81 N
800 kg = 7848 N
Since the rod is half bronze and half steel
800 kg = 7848/2
= 3924 N
Pₙ = Fₙ = 3924 N [n = Bronze]
Pₓ = 3924 N [x = steel]
Given,
σₙ = 90MPa
σₓ = 120MPa
Aₙ = ?
Aₓ = ?
Aₙ = Pₙ/σₙ
Aₙ = 3924/90
Aₙ = 43.6 mm²
Aₓ = Pₓ/σₓ
Aₓ = 3924/120
Aₓ = 32.7 mm²
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