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Harlamova29_29 [7]
3 years ago
7

(a) For a given material, would you expect the surface energy to be greater than, the same as, or less than the grain boundary e

nergy? Why? (b) The grain boundary energy of a small-angle grain boundary is less than for a high-angle one. Why is this so?
Engineering
1 answer:
aksik [14]3 years ago
5 0

Answer:

(a) Surface energy is greater than grain boundary energy due to the fact that the bonds of the atoms on the surface are lower than those of the atoms at the grain boundary. The energy is also directly proportional to the number of bonds created.

(b) The energy of a high-angle grain boundary is higher than that of a small-angle grain boundary because the high-angle grain boundary has a higher misalignment and smaller number of bonds than a small-angle grain boundary.

Explanation:

(a) Surface energy is greater than grain boundary energy due to the fact that the bonds of the atoms on the surface are lower than those of the atoms at the grain boundary. The energy is also directly proportional to the number of bonds created.

(b) The energy of a high-angle grain boundary is higher than that of a small-angle grain boundary because the high-angle grain boundary has a higher misalignment and smaller number of bonds than a small-angle grain boundary.

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Sophie [7]

Answer:

An emergency kit

Explanation:

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2 years ago
A controller on an electronic arcade game consists of a variable resistor connected across the plates of a 0.227 μF capacitor. T
anyanavicka [17]

Answer:

R min = 28.173 ohm

R max = 1.55 × 10^{4}  ohm

Explanation:

given data

capacitor = 0.227 μF

charged to 5.03 V

potential difference across the plates =  0.833 V

handled effectively = 11.5 μs to 6.33 ms

solution

we know that resistance range of the resistor is express as

V(t) = V_o \times e^{t\RC}    ...........1

so R will be

R = \frac{t}{C\times ln(\frac{V_o}{V})}    ....................2

put here value

so for t min 11.5 μs

R = \frac{11.5}{0.227\times ln(\frac{5.03}{0.833})}

R min = 28.173 ohm

and

for t max 6.33 ms

R max = \frac{6.33}{11.5} \times 28.173  

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4 0
3 years ago
A bakery wants to determine how many trays of doughnuts it should prepare each day. Demand is normal with a mean of 15 trays and
shutvik [7]

The number of trays that should be prepared if the owner wants a service level of at least 95% is; 7 trays

<h3>How to utilize z-score statistics?</h3>

We are given;

Mean; μ = 15

Standard Deviation; σ = 5

We are told that the distribution of demand score is a bell shaped distribution that is a normal distribution.

Formula for z-score is;

z = (x' - μ)/σ

We want to find the value of x such that the probability is 0.95;

P(X > x) = P(z > (x - 15)/5) = 0.95

⇒ 1 -  P(z ≤ (x - 15)/5) = 0.95

Thus;

P(z ≤ (x - 15)/5) = 1 - 0.95

P(z ≤ (x - 15)/5) = 0.05

The value of z from the z-table of 0.05 is -1.645

Thus;

(x - 15)/5 = -1.645

x ≈ 7

Complete Question is;

A bakery wants to determine how many trays of doughnuts it should prepare each day. Demand is normal with a mean of 15 trays and standard deviation of 5 trays. If the owner wants a service level of at least 95%, how many trays should he prepare (rounded to the nearest whole tray)? Assume doughnuts have no salvage value after the day is complete. 6 5 4 7 unable to determine with the above information.

Read more about Z-score at; brainly.com/question/25638875

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