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3 years ago

Layout the logical structure of the report.along with headings and subheadings

Engineering

1 answer:

Iteru [2.4K]3 years ago

5 0

Answer:

The typical structure of a report, as shown on this page, is often referred to as IMRAD, which is short for Introduction, Method, Results And Discussion. As reports often begin with an Abstract, the structure may also be referred to as AIMRAD.

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How many ase certifications are there for automotive technicians?

romanna [79]

Answer:

There are 50 ASE certification tests, covering almost every imaginable aspect of the automotive repair and service industry.

Explanation:

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3 years ago

You should be extra careful during the hours of sunrise, sunset, and nighttime because _____. A. of increased law enforcement ac

torisob [31]

Answer:

B, its the only valid answer

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4 years ago

To purchase a new car, you borrow $20,000. The bank offers a 6-year loan at an interest rate of 3.25% compounded annually. If yo

Natalka [10]

Answer:

SPCA factor

Single payment compound amount factor.

Total amount pay A = $24,230.95 (Approx)

Interest paid = $4,230.95 (Approx)

Explanation:

Given:

P = $20,000

n = 6 year

r = 3.25%

Find:

Total amount pay A

Computation:

A=p(1+r)ⁿ

A=20,000[1+3.5%]⁶

A=20,000[(1.0325)⁶]

Total amount pay A = $24,230.95 (Approx)

Interest paid = $24,230.95 - 20,000

Interest paid = $4,230.95 (Approx)

6 0

3 years ago

. In a water cooling tower air enters at a height of 1 m above the ground level with velocity of 20 m/s and leaves the tower at

Zinaida [17]

Answer:

w ( mass flow rate of air ) = 3.16 kg/s

Explanation:

<u>Determine the mass flow rate of air </u>

mass flow rate of water = 1.5 kg/s

Height at which air enters the cooling tower = 1m

velocity of air entering at 1 m = 20 m/s

Height at which air leaves the cooling tower = 7 m

attached below is a detailed solution of the problem

4 0

3 years ago

In a shear box test on sand a shearing force of 800 psf was applied with normal stress of 1750 psf. Find the major and minor pri

ryzh [129]

Answer:

The major and minor stresses are as 2060.59 psf, -310.59 psf and 1185.59 psf.

Explanation:

The major and minor principal stresses are given as follows:

$\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}$

$\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}$

Here

$\sigma_x$ is the normal stress which is 1750 psf
$\sigma_y$ is 0
$\tau_{xy}$ is the shear stress which is 800 psf

So the formula becomes

$\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{max}=\dfrac{1750+0}{2}+\sqrt{\left(\dfrac{1750-0}{2}\right)^2+(800)^2}\\\sigma_{max}=875+\sqrt{\left(875)^2+(800)^2} \\\sigma_{max}=875+\sqrt{765625+640000}\\\sigma_{max}=875+1185.59\\\sigma_{max}=2060.59 \text{psf}$

Similarly, the minimum normal stress is given as

$\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{min}=\dfrac{1700+0}{2}-\sqrt{\left(\dfrac{1700-0}{2}\right)^2+(800)^2}\\\sigma_{min}=875-\sqrt{(875)^2+(800)^2}\\\sigma_{min}=875-\sqrt{765625+640000}\\\sigma_{min}=875-1185.59\\\sigma_{min}=-310.59 \text{ psf}$

The maximum shear stress is given as

$\tau_{max}=\dfrac{\sigma_{max}-\sigma_{min}}{2}\\\tau_{max}=\dfrac{2060.59-(-310.59)}{2}\\\tau_{max}=\dfrac{2371.18}{2}\\\tau_{max}=1185.59 \text{psf}$

5 0

3 years ago