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3 years ago

7

): drivers must slow down from 60 to 40 mi/hr to negotiate a severe curve. A warning sign is visible for a distance of 120 ft. H

ow far in advance of the curve must the sign be located.

1 answer:

Yakvenalex [24]3 years ago

8 0

Answer:

Explanation:

Total Stopping distance is the sum of the reaction distance and the braking distance, such that ;

d=dr + db =1.47St +[ ( Si 2- Sf2 ) / 30(F/0.01G) ]

In this case, the reaction time, t, is the AASHTO standard, or 2. 5 s. The friction factor, F, is based upon the standard AASHTO deceleration rate of 11.2 ft/s2 (F = 11.2/32.2 = 0.348) and the speed is given as 40 mph.

d = 1.47x60x2.5+ [ (60^2-40^2) / 30x 0.348 ]

d = 220.5 + ( (3600-1600) / 10.44 )

d = 220.5 + 191.57

d = 412.07ft.

Since the sign can be seen clearly at 120 ft.

Then the position of the sign should be,

= 412.07 - 120

= 292.07 ft

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An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate (ϵ˙) is found to be

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Answer:

Activation energy for creep in this temperature range is Q = 252.2 kJ/mol

Explanation:

To calculate the creep rate at a particular temperature

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Creep rate at 800⁰C, $\zeta_{800} = C \exp(\frac{-Q}{R (800+273)} )$

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Creep rate at 700⁰C

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$5.5 * 10^{-4} = C \exp(\frac{-Q}{1073R} )$ .................(2)

Divide equation (1) by equation (2)

$\frac{0.01}{5.5 * 10^{-4} } = \exp[\frac{-Q}{1073R} -\frac{-Q}{973R} ]\\18.182= \exp[\frac{-Q}{1073R} +\frac{Q}{973R} ]\\R = 8.314\\18.182= \exp[\frac{-Q}{1073*8.314} +\frac{Q}{973*8.314} ]\\18.182= \exp[0.0000115 Q]\\$

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3 years ago

Why does the compression-refrigeration cycle have a high-pressure side and a low-pressure side?

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Answer: D

Explanation:

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2 years ago

For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration will raise the carbon concentratio

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This question is incomplete, the complete question is;

For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration at Temperature T1 will raise the carbon concentration to 0.44 wt% at a point 1.8 mm from the surface. A separate experiment is performed at T2 that doubles the diffusion coefficient for carbon in steel.

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Answer:

the required time to achieve the same concentration at a 4.9 is 83.733 hrs

Explanation:

Given the data in the question;

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Carbon concentration = 0.444 wt%

thickness at surface x₁ = 1.8 mm = 0.0018 m

thickness at identical steel x₂ = 4.9 mm = 0.0049 m

Now, Using Fick's second law inform of diffusion

$x^2$ / Dt = constant

where D is constant

then

$x^2$ / t = constant

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$x^2_1$ t₂ = t₁ $x^2_2$

t₂ = t₁ $x^2_2$ / $x^2_1$

t₂ = ( $x^2_2$ / $x^2_1$ )t₁

t₂ = $($ $x_2$ / $x_1$ $)^2$ × t₁

so we substitute

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Therefore, the required time to achieve the same concentration at a 4.9 is 83.733 hrs

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3 years ago