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3 years ago

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Suppose that all the dislocations in 3000 mm3 of crystal were somehow removed and linked end to end. Given 1 m =0.0006214 mile,

how far (in miles) would this chain extend for dislocation densities of (a) 10^4 mm^-2 (undeformed metal)? (b) 10^10 mm-2 (cold-worked metal)?

1 answer:

zalisa [80]3 years ago

6 0

Answer:

please let me know the answer to this because I need to know as well. thanks

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Answer:

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Explanation:

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Resistance to impact is an example of a(n)

Anika [276]

Answer:

Mechanical property

Explanation:

MECHANICAL PROPERTIES can be defined as the ability of a metal or material to remain undamaged after different type of forces has been applied or used on them because forces or loads are often applied to metal, material or physical properties which is why MECHANICAL PROPERTIES enables us to know the strength , toughness as well as the hardness of metal and the way this metal perform or react when different forces are applied on them.

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3 years ago

Write the chemical equation of the polymerization reaction to synthesize PS

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Explanation:

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The equation is shown below as:

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3 years ago

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3 years ago

Car B is traveling a distance dd ahead of car A. Both cars are traveling at 60 ft/s when the driver of B suddenly applies the br

vagabundo [1.1K]

Answer:

Explanation:

Using the kinematics equation $v = v_o + a_ct$ to determine the velocity of car B.

where;

$v_o =$ initial velocity

$a_c$ = constant deceleration

Assuming the constant deceleration is = -12 ft/s^2

Also, the kinematic equation that relates to the distance with the time is:

$S = d + v_ot + \dfrac{1}{2}at^2$

Then:

$v_B = 60-12t$

The distance traveled by car B in the given time (t) is expressed as:

$S_B = d + 60 t - \dfrac{1}{2}(12t^2)$

For car A, the needed time (t) to come to rest is:

$v_A = 60 - 18(t-0.75)$

Also, the distance traveled by car A in the given time (t) is expressed as:

$S_A = 60 * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2$

Relating both velocities:

$v_B = v_A$

$60-12t = 60 - 18(t-0.75)$

$60-12t =73.5 - 18t$

$60- 73.5 = - 18t+ 12t$

$-13.5 =-6t$

t = 2.25 s

At t = 2.25s, the required minimum distance can be estimated by equating both distances traveled by both cars

i.e.

$S_B = S_A$

$d + 60 t - \dfrac{1}{2}(12t^2) = 60 * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2$

$d + 60 (2.25) - \dfrac{1}{2}(12*(2.25)^2) = 60 * 0.75 +60((2.25)-0.75) -\dfrac{1}{2}*18*((2.25)-0.750)^2$

d + 104.625 = 114.75

d = 114.75 - 104.625

d = 10.125 ft

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3 years ago