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3 years ago

Lets try to get to 100 sub before charismas day Jordan Gracia 32 sub and 5 videos

Engineering

2 answers:

Natasha_Volkova [10]3 years ago

7 0

Keep on posting this and maybe even more people will subscribe to you :)
and I hope you do achieve it

dedylja [7]3 years ago

4 0

Answer:

make this the brainliest and ill do it

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According to the rules of dimension how should written notes appear

Arturiano [62]

Explanation:

Leaders for notes should be straight, not curved, and point to the center of circular views of holes wherever possible. Leaders should slope at 45°, 30° or 60° with horizontal but may be made at any convenient angle except vertical or horizontal.

7 0

3 years ago

A non-inductive load takes a current of 15A at 125V. An inductor is then connected in series in order that the same current shal

Norma-Jean [14]

Answer:

The inductance of the inductor is 0.051H

Explanation:

From Ohm's law;

V = IR .................. 1

The inductor has its internal resistance referred to as the inductive reactance, X $_{L}$ , which is the resistance to the flow of current through the inductor.

From equation 1;

V = IX $_{L}$

X $_{L}$ = $\frac{V}{I}$ ................ 2

Given that; V = 240V, f = 50Hz, $\pi$ = $\frac{22}{7}$ , I = 15A, so that;

From equation 2,

X $_{L}$ = $\frac{240}{15}$

= 16Ω

To determine the inductance of the inductor,

X $_{L}$ = 2 $\pi$ fL

L = $\frac{X_{L} }{2 \pi f}$

= $\frac{16}{2*\frac{22}{7}*50 }$

= 0.05091

The inductance of the inductor is 0.051H.

4 0

4 years ago

How a single force is resolved along the perpendicular axis acting an angle theta with horizontal?

Nataly [62]

Answer:

A single force, which is acting at angle θ from a horizontal axis, can be resolved into components which act along the perpendicular axis.

Consider the perpendicular axis x and y, where x represents the horizontal axis and y represents vertical axis.

The Force is resolved into 2 parts, one acts along x-axis and is represent by X. The other acts along y-axis and is represented by Y.

From the diagram we can see that the Force and its components X and Y makes up a right angles triangle, where θ is the angle from the x-axis

We know that:

cosθ = Base/Hypotenuse

cosθ = X/F

X = Fcosθ

We know that:

sinθ = Perpendicular/Hypotenuse

sinθ = Y/F

Y = Fsinθ

<h3>Relation of Force and its Components:</h3>

Force F can be represent by:

F = Fcosθ (along x-axis) + Fsinθ (along y-axis)

As they form a right angled triangle, we can use Pythagoras Theorem to show the relation between Force and its components.

Hypotenuse² = Base² + Perpendicular²

F² = X² + Y²

F² = (Fcosθ)² + (Fsinθ)²

$F = \sqrt{(F\cos\theta)^2+(Fsin\theta)^2}$

Where θ can be found by using any of the trignometric functions.

6 0

3 years ago

Two added to four times a number, minus 3 times the number, equals 5.

vladimir1956 [14]

<h2>Answer:</h2>

<u>x= 3</u>.

<h2>Explanation:</h2>

<em>What is presented in this problem is basically an equation in verbal form.</em>

<em />

<h3>1. Write the equation.</h3>

$2+4x-3x=5$

<h3>2. Solve for x.</h3>

$2+4x-3x=5\\ \\2+x=5\\ \\x=5-2\\ \\x=3$

<h3>3. Express the result.</h3>

x= 3.

8 0

2 years ago

Air modeled as an ideal gas enters a combustion chamber at 20 lbf/in.2

motikmotik

Answer:

The answer is " $112.97 \ \frac{ft}{s}$ "

Explanation:

Air flowing into the $p_1 = 20 \ \frac{lbf}{in^2}$

Flow rate of the mass $m = 230.556 \frac{lbm}{s}$

inlet temperature $T_1 = 700^{\circ} F$

Pipeline $A= 5 \times 4 \ ft$

Its air is modelled as an ideal gas Apply the ideum gas rule to the air to calcule the basic volume v:

$\to \bar{R} = 1545 \ ft \frac{lbf}{lbmol ^{\circ} R}\\\\ \to M= 28.97 \frac{lb}{\bmol}\\\\ \to pv=RT \\\\\to v= \frac{\frac{\bar{R}}{M}T}{p}$

$= \frac{\frac{1545}{28.97}(70^{\circ}F+459.67)}{20} \times \frac{1}{144}\\\\=9.8 \frac{ft3}{lb}$

$V= \frac{mv}{A}$

$= \frac{230.556 \frac{lbm}{s} \times 9.8 \frac{ft^3}{lb}}{5 \times 4 \ ft^2}\\\\= 112.97 \frac{ft}{s}$

8 0

3 years ago