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2 years ago

Can someone help me LA project pls :((

Engineering

1 answer:

sweet [91]2 years ago

4 0

The three-step method that can be used to find the topic of an article are:

Pre-writing,
drafting
final revising stage

When making a review of an article and you want to find the topic, it is advised that you:

Read and understand the article
Write out the main points
With this you can find the theme
Then you can create a suitable title

<h3>What is a Text Review?</h3>

This refers to the academic process that is done on a piece of writing that aims to critique it.

Hence, we can see that the aforementioned steps would be beneficial to you.

You might be interested in

Define a function pyramid_volume with parameters base_length, base_width, and pyramid_height, that returns the volume of a pyram

Maslowich

Hi, you haven't provided the programing language in which you need the code, I'll just explain how to do it using Python, and you can apply a similar method for any programming language.

Answer:

1. def pyramid_volume(base_length, base_width, pyramid_height):

2. volume = base_length*base_width*pyramid_height/3

3. return(volume)

Explanation step by step:

In the first line of code, we define the function pyramid_volume and it's input parameters
In the second line, we perform operations with the input values to get the volume of the pyramid with a rectangular base, the formula is V = l*w*h/3
In the last line of code, we return the volume

In the image below you can see the result of calling the function with input 4.5, 2.1, 3.0.

5 0

3 years ago

It is given that 50 kg/sec of air at 288.2k is iesntropically compressed from 1 to 12 atm. Assuming a calorically perfect gas, d

denis23 [38]

The exit temperature is 586.18K and compressor input power is 14973.53kW

Data;

Mass = 50kg/s
T = 288.2K
P1 = 1atm
P2 = 12 atm

<h3>Exit Temperature </h3>

The exit temperature of the gas can be calculated isentropically as

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y}\\ y = 1.4\\ C_p= 1.005 Kj/kg.K\\$

Let's substitute the values into the formula

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y} \\\frac{T_2}{288.2} = (\frac{12}{1})^\frac{1.4-1}{1.4} \\ T_2 = 586.18K$

The exit temperature is 586.18K

<h3>The Compressor input power</h3>

The compressor input power is calculated as

$P= mC_p(T_2-T_1)\\P = 50*1.005*(586.18-288.2)\\P= 14973.53kW$

The compressor input power is 14973.53kW

Learn more on exit temperature and compressor input power here;

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6 0

2 years ago

What are the 3 dimensions that used in isometric sketches?

noname [10]

Answer:

The three dimensions shown in an isometric drawing are the height, H, the length, L, and the depth, D

Explanation:

An isometric drawing of an object in presents a pictorial projection of the object in which the three dimension, views of the object's height, length, and depth, are combined in one view such that the dimensions of the isometric projection drawing are accurate and can be measured (by proportion of scale) to draw the different views of the object or by scaling, for actual construction of the object.

5 0

3 years ago

. Two rods, with masses MA and MB having a coefficient of restitution, e, move

GarryVolchara [31]

Answer:

a) $V_A = \frac{(M_A - eM_B)U_A + M_BU_B(1+e)}{M_A + M_B}$

$V_B = \frac{M_AU_A(1+e) + (M_B - eM_A)U_B}{M_A + M_B}$

b) $U_A = 3.66 m/s$

$V_B = 4.32 m/s$

c) Impulse = 0 kg m/s²

d) percent decrease in kinetic energy = 47.85%

Explanation:

Let $U_A$ be the initial velocity of rod A

Let $U_B$ be the initial velocity of rod B

Let $V_A$ be the final velocity of rod A

Let $V_B$ be the final velocity of rod B

Using the principle of conservation of momentum:

$M_AU_A + M_BU_B = M_AV_A + M_BV_B$ ............(1)

Coefficient of restitution, $e = \frac{V_B - V_A}{U_A - U_B}$

$V_A = V_B - e(U_A - U_B)$ ........................(2)

Substitute equation (2) into equation (1)

$M_AU_A + M_BU_B = M_A(V_B - e(U_A - U_B)) + M_BV_B$ ..............(3)

Solving for $V_B$ in equation (3) above:

$V_B = \frac{M_AU_A(1+e) + (M_B - eM_A)U_B}{M_A + M_B}$ ....................(4)

From equation (2):

$V_B = V_A + e(U_A -U_B)$ ......(5)

Substitute equation (5) into (1)

$M_AU_A + M_BU_B = M_AV_A + M_B(V_A + e(U_A -U_B))$ ..........(6)

Solving for $V_A$ in equation (6) above:

$V_A = \frac{(M_A - eM_B)U_A + M_BU_B(1+e)}{M_A + M_B}$ .........(7)

$M_A = 2 kg\\M_B = 1 kg\\U_B = -3 m/s( negative x-axis)\\e = 0.65\\U_A = ?$

Rod A is said to be at rest after the impact, $V_A = 0 m/s$

Substitute these parameters into equation (7)

$0 = \frac{(2 - 0.65*1)U_A - (1*3)(1+0.65)}{2+1}\\U_A = 3.66 m/s$

To calculate the final velocity, $V_B$ , substitute the given parameters into (4):

$V_B = \frac{(2*3.66)(1+0.65) - (1 - (0.65*2))*3}{2+1}\\V_B = 4.32 m/s$

c) Impulse, $I = M_AV_A + M_BV_B - (M_AU_A + M_BU_B)$

$I = (2*0) + (1*4.32) - ((2*3.66) + (1*-3))$

I = 0 $kg m/s^2$

d) % $\triangle KE = \frac{(0.5 M_A V_A^2 + 0.5 M_B V_B^2) - ( 0.5 M_A U_A^2 + 0.5 M_B U_B^2)}{0.5 M_A U_A^2 + 0.5 M_B U_B^2} * 100\%$

% $\triangle KE = \frac{((0.5*2*0) + (0.5 *1*4.32^2)) - ( (0.5 *2*3.66^2) + 0.5*1*(-3)^2))}{ (0.5 *2*3.66^2) + 0.5*1*(-3)^2)} * 100\%$

% $\triangle KE = -47.85 \%$

7 0

3 years ago

If a worker currently makes $425.00 per week, what Is the gross amount they’ll earn if they work every week of the year

olga nikolaevna [1]

Answer:

we know that

A worker currently makes \$425.00 per week

remember that

1\ year=52\ weeks

by proportion find the amount that the worker will earn in one year

\frac{425}{1} \frac{\$}{week} =\frac{x}{52} \frac{\$}{weeks} \\ \\x=52*425 \\ \\x=\$22,100

therefore

the answer is

\$22,100

Explanation:

4 0

3 years ago