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3 years ago

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A high-voltage direct-current (dc) transmission line between Celilo, Oregon and Sylmar, California is 845 mi in length. The line

contains four conductors, each weighing 2460 lbper 1000 ft.How many kilograms of conductor are in the line?

1 answer:

Gennadij [26K]3 years ago

4 0

Answer:

<u>Total Mass of Conductors = 19913661.3 kg</u>

Explanation:

The total length is given as:

Total Length = 845 mi

Converting it to ft:

Total length = (845 mi)(5280 ft/1 mi)

Total length = 4,461,600 ft

The weight of one conductor per unit length is:

Mass of one conductor per unit length = 2460 lb/1000 ft

Mass of one conductor per unit length = 2.46 lb/ft

Since there are 4 conductors, therefore, total weight of all conductors per unit length will be:

Mass of all four conductors per unit length = 4 x 2.46 lb/ft

Mass of all four conductors per unit length = (9.84 lb/ft)(0.453592 kg/1 lb)

Mass of all four conductors per unit length = 4.463 kg/ft

Therefore, total mass of conductors in the line will be:

Total Mass of Conductors = (Mass of all four conductors per unit length)(Total Length)

Total Mass of Conductors = (4.463 kg/ft)(4,461,600 ft)

<u>Total Mass of Conductors = 19913661.3 kg</u>

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A train starts from rest at station A and accelerates at 0.6 m/s^2 for 60 s. Afterwards it travels with a constant velocity for

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Answer:

The distance between the station A and B will be:

$x_{A-B}=55.620\: km$

Explanation:

Let's find the distance that the train traveled during 60 seconds.

$x_{1}=x_{0}+v_{0}t+0.5at^{2}$

We know that starts from rest (v(0)=0) and the acceleration is 0.6 m/s², so the distance will be:

$x_{1}=\frac{1}{2}(0.6)(60)^{2}$

$x_{1}=1080\: m$

Now, we need to find the distance after 25 min at a constant speed. To get it, we need to find the speed at the end of the first distance.

$v_{1}=v_{0}+at$

$v_{1}=(0.6)(60)=36\: m/s$

Then the second distance will be:

$x_{2}=v_{1}*1500$

$x_{2}=(36)(1500)=54000\: m$

The final distance is calculated whit the decelerate value:

$v_{f}^{2}=v_{1}^{2}-2ax_{3}$

The final velocity is zero because it rests at station B. The initial velocity will be v(1).

$0=36^{2}-2(1.2)x_{3}$

$x_{3}=\frac{36^{2}}{2(1.2)}$

$x_{3}=540\: m$

Therefore, the distance between the station A and B will be:

$x_{A-B}=x_{1}+x_{2}+x_{3}$

$x_{A-B}=1080+54000+540=55.620\: km$

I hope it helps you!

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Suppose that you can throw a projectile at a large enough v0 so that it can hit a target a distance R downrange. Given that you

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Answer:

$\theta_1=15^o\\\theta_2=75^o$

Explanation:

<u>Projectile Motion</u>

In projectile motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration (assuming no friction), and the acceleration in the vertical direction is always the acceleration of gravity. The basic formulas are shown below:

$V_x=V_{ox}=V_ocos\theta$

Where $\theta$ is the angle of launch respect to the positive horizontal direction and Vo is the initial speed.

$V_y=V_{oy}-gt=V_osin\theta-gt$

The horizontal and vertical distances are, respectively:

$x=V_{o}cos\theta t$

$\displaystyle y=y_o+V_{o}sin\theta t-\frac{gt^2}{2}$

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$\displaystyle t_f=\frac{2V_osin\theta}{g}$

Using this time in the horizontal distance, we find the Range or maximum horizontal distance:

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This is the general expression to determine the angles at which the projectile can be launched to hit the target. Recall the angle can have to values for fixed positive values of its sine:

$\displaystyle \theta_1=\frac{asin\left(\frac{R.g}{V_o^2}\right)}{2}$

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$\displaystyle \theta_1=\frac{asin\left(\frac{70\times 9.8}{37^2}\right)}{2}$

$\theta_1=15^o$

And

$\theta_2=90^o-15^o=75^o$

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