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2 years ago

The Texas Sure program is designed to:

1 answer:

4 0

TexasSure is designed to reduce the number of uninsured drivers and cut costs for responsible Texans, who now pay almost $900 million a year to protect themselves against those with no coverage. Currently, an estimated 20 percent of Texas drivers are uninsured.

TexasSure is the financial responsibility verification program created by the 79th Texas Legislature, Regular Session, in Senate Bill 1670.

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A medium-sized jet has a 3.8-mm-diameter fuselage and a loaded mass of 85,000 kg. The drag on an airplane is primarily due to th

SCORPION-xisa [38]

Answer:

$F_{thrust}$ ≅ 111 KN

Explanation:

Given that;

A medium-sized jet has a 3.8-mm-diameter i.e diameter (d) = 3.8

mass = 85,000 kg

drag co-efficient (C) = 0.37

(velocity (v)= 230 m/s

density (ρ) = 1.0 kg/m³

To calculate the thrust; we need to determine the relation of the drag force; which is given as:

$F_{drag}$ = $\frac{1}{2}$ × CρAv²

where;

ρ = density of air wind.

C = drag co-efficient

A = Area of the jet

v = velocity of the jet

From the question, we can deduce that the jet is in motion with a constant speed; as such: the net force acting on the jet in the air = 0

SO, $F_{drag}-F_{thrust} = 0$

We can as well say:

$F_{drag}= F_{thrust}$

We can now replace $F_{thrust} with F_{drag}$ in the above equation.

Therefore, $F_{thrust}$ = $\frac{1}{2}$ × CρAv²

The A which stands as the area of the jet is given by the formula:

$A=\frac{\pi d^2}{4}$

We can now have a new equation after substituting our A into the previous equation as:

$F_{thrust}$ = $\frac{1}{2}$ × Cρ $(\frac{\pi d^2}{4})v^2$

Substituting our data from above; we have:

$F_{thrust}$ = $\frac{1}{2}$ × $(0.37)(1.0kg/m^3)(\frac{\pi(3.8m)^2 }{4})(230m/s)^2$

$F_{thrust}$ = $\frac{1}{8} (0.37)(1.0kg/m^3)({\pi(3.8m)^2 })(230m/s)^2$

$F_{thrust}$ = 110,990N

$F_{thrust}$ in N (newton) to KN (kilo-newton) will be:

$F_{thrust}$ = $(110,990N)*\frac{1KN}{1,000N}$

$F_{thrust}$ = 110.990 KN

$F_{thrust}$ ≅ 111 KN

In conclusion, the jet engine needed to provide 111 KN thrust in order to cruise at 230 m/s at an altitude where the air density is 1.0 kg/m³.

5 0

3 years ago

What is the average linear (seepage) velocity of water in an aquifer with a hydraulic conductivity of 6.9 x 10-4 m/s and porosit

jeka94

Answer:

a. 0.28

Explanation:

Given that

porosity =30%

hydraulic gradient = 0.0014

hydraulic conductivity = 6.9 x 10⁻4 m/s

We know that average linear velocity given as

$v=\dfrac{K}{n_e}\dfrac{dh}{dl}$

$v=\dfrac{6.9\times 10^{-4}}{0.3}\times0.0014\ m/s$

$v=3.22\times 10^{-6}\ m/s$

The velocity in m/d ( 1 m/s =86400 m/d)

v= 0.27 m/d

So the nearest answer is 'a'.

a. 0.28

4 0

3 years ago

Comparación de hipotecas Los Chos

aleksklad [387]

Answer:

I don't understand the language French sorry can't answer

3 0

3 years ago

Brainstorming is the problem-solving method engineers use most.<br>True<br>False

Lilit [14]

Answer:

I'm pretty sure it's false

Explanation:

Brainstorm is part of a problem-solving method. you can't solve a problem with nothing but brainstorming

7 0

3 years ago

A steady, incompressible, two-dimensional velocity field is given by the following components in the x-y plane: u=1.85+2.05x+0.6

amm1812

1) $a_x=4.287+2.772x\\a_y=-5.579+2.772y$

2) 8.418

Explanation:

The two components of the velocity field in x and y for the field in this problem are:

$u=1.85+2.05x+0.656y$

$v=0.754-2.18x-2.05y$

The x-component and y-component of the acceleration field can be found using the following equations:

$a_x=\frac{du}{dt}+u\frac{du}{dx}+v\frac{du}{dy}$

$a_y=\frac{dv}{dt}+u\frac{dv}{dx}+v\frac{dv}{dy}$

The derivatives in this problem are:

$\frac{du}{dt}=0$

$\frac{dv}{dt}=0$

$\frac{du}{dx}=2.05$

$\frac{du}{dy}=0.656$

$\frac{dv}{dx}=-2.18$

$\frac{dv}{dy}=-2.05$

Substituting, we find:

$a_x=0+(1.85+2.05x+0.656y)(2.05)+(0.754-2.18x-2.05y)(0.656)=\\a_x=4.287+2.772x$

And

$a_y=0+(1.85+2.05x+0.656y)(-2.18)+(0.754-2.18x-2.05y)(-2.05)=\\a_y=-5.579+2.772y$

In this part of the problem, we want to find the acceleration at the point

(x,y) = (-1,5)

So we have

x = -1

y = 5

First of all, we substitute these values of x and y into the expression for the components of the acceleration field:

$a_x=4.287+2.772x\\a_y=-5.579+2.772y$

And so we find:

$a_x=4.287+2.772(-1)=1.515\\a_y=-5.579+2.772(5)=8.281$

And finally, we find the magnitude of the acceleration simply by applying Pythagorean's theorem:

$a=\sqrt{a_x^2+a_y^2}=\sqrt{1.515^2+8.281^2}=8.418$

4 0

3 years ago