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3 years ago

How many paragraphs are good for an essay

Engineering

2 answers:

zhannawk [14.2K]3 years ago

3 0

Answer:

It depends what type of essay you're writing. Normally you'll want an intro, three body paragraphs and then a conclusion, but if you're writing an argumentative essay you will want an intro par, three body par, a counterclaim and rebuttal par, and a conclusion par.

Explanation:

Typically 5 or 6 paragraphs will work, really just depends what type of essay.

NISA [10]3 years ago

3 0

Answer:

In its simplest form, an essay can consist of three paragraphs with one paragraph being devoted to each section. Proponents of the five paragraph essay say that the body text should consist of three paragraphs, but in reality, it's fine to write more or fewer paragraphs in this section. An essay is split into sections not paragraphs. While shorter essays will often have one paragraph per section most will have multiple. Paragraph and section are not interchangeable terms. If you wish to only have two paragraphs then you will need to make the core point sections into one paragraph. An essay can be pretty much any length, as long as it's relatively short. Some essays are just one paragraph, and some are several pages long. As long as you say what needs to be said, you can write four paragraphs, six paragraphs, or thirty paragraphs. 1500 words is 8 to 15 paragraphs for essays, 15 to 30 paragraphs for easy writing.

Explanation:

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Info security:

il63 [147K]

Answer:

True

Explanation:

Dual home host - it is referred to as the firewall that is incorporated with two or more networks. out of these two networks, one is assigned to the internal network and the other is for the network. The main purpose of the dual-homed host is to ensure that no Internet protocol traffic is induced between both the network.

The most simple example of a dual-homed host is a computing motherboard that is provided with two network interfaces.

7 0

4 years ago

What is the velocity of flow in an asphalt channel that has a hydraulic radius of 3.404 m, length of 200 m and bed slope of 0.00

yKpoI14uk [10]

Answer:

The velocity of flow is 10.0 m/s.

Explanation:

We shall use Manning's equation to calculate the velocity of flow

Velocity of flow by manning's equation is given by

$V=\frac{1}{n}R^{2/3}S^{1/2}$

where

n = manning's roughness coefficient

R = hydraulic radius

S = bed slope of the channel

We know that for an asphalt channel value of manning's roughness coefficient = 0.016

Applying values in the above equation we obtain velocity of flow as

$V=\frac{1}{0.016}\times 3.404^{2/3}\times 0.005^{1/2}\\\\\therefore V=10.000m/s$

7 0

3 years ago

A horizontal 2-m-diameter conduit is half filled with a liquid (SG=1.6 ) and is capped at both ends with plane vertical surfaces

luda_lava [24]

Answer:

Resultant force = 639 kN and it acts at 0.99m from the bottom of the conduit

Explanation:

The pressure is given as 200 KPa and the specific gravity of the liquid is 1.6.

The resultant force acting on the vertical plate, Ft, is equivalent to the sum of the resultant force as a result of pressurized air and resultant force due to oil, which will be taken as F1 and F2 respectively.

Therefore,

Ft = F1 + F2

According to Pascal's law which states that a change in pressure at any point in a confined incompressible fluid is transmitted throughout the fluid such that the same change occurs everywhere, the air pressure will act on the whole cap surface.

To get F1,

F1 = p x A

= p x (πr²)

Substituting values,

F1 = 200 x π x 1²

F1 = 628.32 kN

This resultant force acts at the center of the plate.

To get F2,

F2 = Π x hc x A

F2 = Π x (4r/3π) x (πr²/2)

Π - weight density of oil,

A - area on which oil pressure is acting,

hc - the distance between the axis of the conduit and the centroid of the semicircular area

Π = Specific gravity x 9.81 x 1000

Therefore

F2 = 1.6 x 9.81 x 1000 x (4(1)/3π) x (π(1)²/2)

F2 = 10.464 kN

Ft = F1 + F2

Ft = 628.32 + 10.464

Ft = 638.784 kN

The resultant force on the surface is 639 kN

Taking moments of the forces F1 and F2 about the centre,

Mo = Ft x y

Ft x y = (F1 x r) + F2(1 - 4r/3π)

Making y the subject,

y = (628.32 + 10.464(1 - 4/3π)/ 638.784

y = 0.993m

7 0

4 years ago

Give me a smart goals to achieve

S_A_V [24]

Just a suggestion: i would start with short-term goals to help you build up to your long-term goals

7 0

3 years ago

Read 2 more answers

A large steel tower is to be supported by a series of steel wires; it is estimated that the load on each wire will be 19,000N. D

zhuklara [117]

Answer:

11.6 mm

Explanation:

With a factor of safety of 5 and a yield strength of 900 MPa the admissible stress is:

σadm = strength / fos

σadm = 900 / 5 = 180 MPa

The stress is the load divided by the section:

σ = P / A

σ = 4*P / (π*d^2)

Rearranging:

d^2 = 4*P / (π*σ)

$d = \sqrt{4*P / (\pi*\sigma)}$

$d = \sqrt{4*19000 / (\pi*180*10^6)} = 0.0116 m = 11.6 mm$

7 0

3 years ago