High speed increases the risk of collision because of ALL of these things EXCEPT:

Derive the expression ε=ln(1+e), where ε is the true strain and e is the engineering strain. Note that this expression is not va

Rudiy27

The formula for true strain after derivation from basic terms is; ε_t = In(1 + ε_e)

<h3>How to derive the expression for True Strain?</h3>

Formula for Engineering Stress is;

σ_e = Load/Area

Formula for true stress is;

σ_t = Force/Instantaneous Area

Formula for Engineering Strain is;

ε_e = ΔL/L₀

Formula for true strain is;

dε_t = dL/L

Total true strain is gotten from;

ε_t = ∫(dL/L) between boundaries of L_f and L_o

When we integrate between those boundaries, we have;

ε_t = In[(L₀ + ΔL)/L₀

⇒ ε_t = In[(1+ ΔL/L₀)

⇒ ε_t = In(1 + ε_e)

Read more about True Strain at; brainly.com/question/20717759

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6 0

2 years ago

In a river reach, the rate of inflow at any time is 350 cfs and the rate of outflow is 285 cfs. After 90 min, the inflow and out

Semenov [28]

Answer:

change in storage = -310,500 ft^3

intital storage= 3.67 acre ft

Explanation:

Given data:

Rate of inflow = 350 cfs

Rate of outflow = 285 cfs

After 90 min, rate of inflow = 250 cfs

Rate of outflow = 200 cfs

final storage = 10.8 acre-ft

calculating the average inflow and outflow

average inflow $= \frac{(350+250)}{2} = 300 cfs$

average outlow $= \frac{(285+200)}{2} = 242.5 cfs$

total amount of water drain during the period of one hour

= (average outflow - average inflow) *60*90

= (242.5 - 300)*60*90 = -310,500

change in storage is calculate as

= -310,500 ft^3

in cubic meter

= -310500/35.315 = 8792.30 cm^3

in acre-ft

= -310,500/43560 = 7.13 acre ft

initial storage = 10.8 - 7.13 = 3.67 acre ft

3 0

3 years ago

Given that the skin depth of graphite at 100 (MHz) is 0.16 (mm), determine (a) the conductivity of graphite, and (b) the distanc

Andrei [34K]

Answer:

the answer is below

Explanation:

a) The conductivity of graphite (σ) is calculated using the formula:

$\delta=\frac{1}{\sqrt{\pi f \mu \sigma} }\\\\\sigma =\frac{1}{\pi f \mu \delta^2}$

where f = frequency = 100 MHz, δ = skin depth = 0.16 mm = 0.00016 m, μ = 0.0000012

Substituting:

$\sigma =\frac{1}{\pi *10^6* 0.0000012*0.00016^2}=0.99*10^4\ S/m$

b) f = 1 GHz = 10⁹ Hz.

$\alpha=\sqrt{\pi f \mu \sigma} = \sqrt{0.0000012*10^9*\pi*0.99*10^5}=1.98*10^4\ Np/m\\\\20log_{10} e^{-\alpha z}=-30\ dB\\\\(-\alpha z)log_{10} e=-1.5 \\\\z=\frac{-1.5}{log_{10} e*-\alpha} =1.75*10^{-4}\ m=0.175\ mm$

4 0

3 years ago

A hole of diameter D = 0.25 m is drilled through the center of a solid block of square cross section with w = 1 m on a side. The

attashe74 [19]

Answer:

$q=4.013\:\:kW\\\\T_1=278.91\:\:^{\circ}C\\\\T_2=275.82\:\:^{\circ}C$

Explanation:

$R_{conv,1}=(h_1\pi D_1L)^{-1}=(50*0.25*2)^{-1}=0.01273\:\:K/W\\\\R_{conv,2}=(h^2*4wL)^{-1}=(4*4*1)^{-1}=0.0625\:\:K/W\\\\R_{cond(2D)}=(Sk)^{-1}=(8.59*150)^{-1}=0.00078\:\:K/W$

So, heat rate can be calculated as follows:

$q=\frac{T_{\infty,1}-T_{\infty,2}}{R_{conv,1}+R_{conv,2}+R_{cond(2D)}} =\frac{330-25}{0.076} =4.013\:\:kW$

Surface temperatures can be calculated as follows:

$T_1=T_{\infty,1}-qR_{conv,1}=330-51.09=278.91\:\:^{\circ}C\\\\T_2=T_{\infty,2}+qR_{conv,2}=25+250.82=275.82\:\:^{\circ}C$

6 0

4 years ago

What is the definition of a tolerance on a dimension typically found on technical drawings?

Alinara [238K]

Tolerance is the acceptable amount of dimensional variation that still allows a part to perform as designed.

Any process will have variation and depending on the severity of the function some tolerance will be very small. For example the sheet metal thickness on portion of a space shuttle will have a much tighter tolerance than the thickness of a piece of lumber to build a house. Tighter tolerance of processes typically are related to more process control (e.g. money) thus designs should be fully vetted with process team before placing on a drawing.

7 0

3 years ago