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Daniel [21]
4 years ago
5

Suppose that all the dislocations in 4700 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) 105 mm-2 (undeformed metal)? Enter your answer for part (a) in accordance to the question statement miles (b) 1010 mm-2 (cold-worked metal)?
Physics
1 answer:
san4es73 [151]4 years ago
6 0

Answer:

a. The required chain length = 292.1 miles

b. Length of the required chain = 29.2*10^6miles

Explanation:

Cold work is the work performed on a material at a temperature below the re-crystallization temperature of the material.

Dislocation density is the measure of the total number of dislocation of a crystalline material of unit volume.

Deformation is the process of change in an objects shape due to the application of force

a. In order to find the length of required chain for the given dislocation density

length of required chain = total length of dislocation line =  dislocation density * volume

where dislocation density = 10^5mm^{-2}

and volume is given as = 4700mm^3

=10^5*4700mm(\frac{1mi}{1.609*10^6mm} )=292.1miles

b. To find the length of required chain

length of required chain = dislocation density * volume

where dislocation density = 10^{10}mm^{-2}

we substitute in the equation above to get

=10^{10}mm^{-2}*4700mm^3\\=47*10^{12}mm*(\frac{1mile}{1.609*10^6mm} )=29.2*10^{6}miles

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Explanation:

(a)  Formula to calculate the capacitance is as follows.

           V_{c} = V_{o} (e^(\frac{-t}{RC}))

Now, putting the given values into the above formula as follows.

          V_{c} = V_{o} (e^(\frac{-t}{RC}))

          3.5 V = 12 V (e^(\frac{-3.50 sec}{3.90 \times C}))

    e^(\frac{-3.50 sec}{3.90 \times C}) = 0.291

or,       \frac{-3.50 sec}{3.90 \times C} = ln (0.291)

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Hence, the value of capacitance is 0.729 F.

(b)   Formula to calculate the constant of circuit is as follows.

               T = R \times C

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4 years ago
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Elena L [17]

Answer:

1) 341 Hz

Explanation:

When a string vibrates, it can vibrate with different frequencies, corresponding to different modes of oscillations.

The fundamental frequency is the lowest possible frequency at which the string can vibrate: this occurs when the string oscillate in one segment only.

If the string oscillates in n segments, we say that it is the n-th mode of vibration, or n-th harmonic.

The frequency of the n-th harmonic is given by

f_n = nf_1

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Here we have:

f_3=512 Hz is the frequency of the 3rd harmonic

So the fundamental frequency is

f_1=\frac{f_3}{3}=\frac{512}{3}=170.7 Hz

And so, the frequency of the 2nd harmonic is:

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8) A plastic rod, initially uncharged, is rubbed with wool and obtains a charge of 10 C. What is the charge on the wool after ru
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