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2 years ago

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A cylinder with a 0.25 m inner diameter and a 0.43 m outer diameter is internally pressurized to 175 MPa. a) Determine the maxim

um shear stress at the outer surface of the cylinder. b) Using DET, determine the equivalent stress in the cylinder.

1 answer:

Ivan2 years ago

8 0

The answer is (b) the second one

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Joe Bruin has a big lawn in front of his house that is 30 meters wide and 20 meters long. Josephine makes him go out and mow the

zysi [14]

<u>Explanation:</u>

5 Horsepower for 30 mins,

(5)(745.7) = 3.7285 KW power delivered

General Efficiency of IC engine = 20%

Power required = $\frac{3 \cdot 7285}{0 \cdot 2}=18 \cdot 6425 kw$

Energy required per week,

=P × Time = 18.64 × 60 × 30 = 33.5565 MJ

Lawn area = (30) (20) = 600 $m^{2}$

let sunlight hours be 8 hours

Hence, solar power input on lawn,

=5.62×3600 = 20232 kJ/ $m^{2}$ /day

energy input in lawn = (600) (20232) (7)

= 84974.4 mJ/week

Chemical efficiency by photosynthesis = 4%

Chemical content in grass = (84974.4) (0.04)

= 3398.97 mJ

Mass of the clippers $=(30)(20)(1 \cdot 096)^{2}(667)$

$=478632 \cdot 33$ pounds

Removing water content,

dried grass clippings $\(=95726.46\) pound$

= 11533.25 gallons

Trash cans repaired

$=\frac{11533}{50} =230.66\\=231 cans$

By burning the gas, total energy input = 3398.97 MJ × 0.2

= 679.794 MJ

Efficiency of steeling engine = 20%

Energy output by engine = 679.794 ×0.2

= 135.96 mJ

Energy required by mover = 33.5565 mJ

Hence, Energy (output) ⇒ energy required

5 0

2 years ago

This problem has been solved!

lisov135 [29]

Answer: a) 135642 b) 146253

Explanation:

A)

1- the bankers algorithm tests for safety by simulating the allocation for predetermined maximum possible amounts of all resources, as stated this has the greatest degree of concurrency.

3- reserving all resources in advance helps would happen most likely if the algorithm has been used.

5- Resource ordering comes first before detection of any deadlock

6- Thread action would be rolled back much easily of Resource ordering precedes.

4- restart thread and release all resources if thread needs to wait, this should surely happen before killing the thread

2- only option practicable after thread has been killed.

Bii) ; No. Even if deadlock happens rapidly, the safest sequence have been decided already.

5 0

2 years ago

BCC lithium has a lattice parameter of 3.5089x10^-8cm and contains one vacancy per 200unit cells.Calcuate (a)the number of vacan

professor190 [17]

Answer:

a) The number of vacancies percubic centimeter is $1.1573\times 10^{20} vacancy/cm^3$ .

b) The density of Lithium is $1.076 g/cm^3$ .

Explanation:

Edge length of the lithium unit cell = a = $3.5089\times 10^{-8} cm$

Volume of the unit cell = V

$V=a^3=(3.5089\times 10^{-8} cm)^3=4.3203\times 10^{-23} cm^3$

The lithium crystal has one vacancy per 200unit cells.

$\frac{1}{200} vacancy/unit cell=0.005$ vacancy/unit cell

The number of vacancies percubic centimeter:

$\frac{0.005}{V}=\frac{0.005 vacancy}{4.3203\times 10^{-23} cm^3}$

$=1.1573\times 10^{20} vacancy/cm^3$

The number of vacancies percubic centimeter is $1.1573\times 10^{20} vacancy/cm^3$ .

b)

Atomic mass of lithium = 7 g/mol

Number of atom in a unit cell = z = 2

Volume of the unit cell = V

$\rho =\frac{z\times M}{N_A\times V}$

$=\frac{4\times 7g/mol}{6.022\times 10^{23} mol^{-1}\times 4.3203\times 10^{-23} cm^3}=1.076 g/cm^3$

The density of Lithium is $1.076 g/cm^3$ .

6 0

2 years ago

A non-licensed person may be the SOLE owner of a civil, electrical, or mechanical engineering business under which of the follow

kotegsom [21]

Answer:

(d) None. No provisions exist.

Explanation:

B&P Code § 6738 prohibits a non-licensed person from being the sole proprietor of an engineering business. The non-licensed can be a partner in an engineering business that offers civil, electrical, or mechanical services. It is mandatory that at least one licensed engineer must be a co-owner of the business.

5 0

2 years ago

Suppose a possibly biased die is rolled 30 times and that the face containing

WARRIOR [948]

Answer:

p < α

0.01297 < 0.05

Since the p value is less than the α value therefore, we reject the null hypothesis so we have evidence to conclude that the die is biased.

Explanation:

H₀: The die is not biased

Ha: The die is biased

We can apply binomial distribution and determine whether the die is biased or not. (we can also perform z-test, it will provide similar results)

We know that a binomial distribution is given by

P(x; n, p) = nCx pˣ (1 - p)ⁿ⁻ˣ

Where p is the probability of success and 1 - p is the probability of failure, n is number of trials and x is the variable of interest.

For the given problem,

Total trials are n = 30

When you roll a die, there are total 6 possible outcomes,

The probability of getting the face containing two pips on each trial is

p = 1/6

p = 0.1667

The variable of interest is x = 10

P(10; 30, 0.1667) = ³⁰C₁₀*0.1667¹⁰*(1 - 0.1667)³⁰⁻¹⁰

P(10; 30, 0.1667) = (30045015)*(0.1667)¹⁰*(0.8333)²⁰

P(10; 30, 0.1667) = 0.01297

Assuming that the level of significance is α = 0.05 then

p < α

0.01297 < 0.05

Since the p value is less than the α value therefore, we reject the null hypothesis so we have evidence to conclude that the die is biased.

3 0

2 years ago