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3 years ago

The voltage of electricity traveling away from a power plant is very high. How high may it be? Why is the voltage so high?

1 answer:

8 0

- The voltage travelling away from a power plant through transmission lines is very high, and it is typically of hundreds of kilovolts (typical values are between 138 kV and 765 kV).

- The main reason to use these high values of voltage is to reduce power dissipation.
In fact, the cables that are used to transmit electricity have a certain resistance R which is fixed. The power generated from the power plant and that should be transmitted through the lines is P, and it is also fixed.
The power dissipated through the cables is calculated as
$P_{diss}=I^2 R = (\frac{P}{V} )^2 R$
where I is the current and V the voltage.
As it can be seen, using higher voltages reduce the amount of power dissipated through the lines (while using higher currents will have the opposite effect).

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How many states in the world are there?

aliya0001 [1]

Answer:

There are 193 member states. The UN recognises 206 states altogether.

Explanation:

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3 years ago

About 2% of our solar nebula consisted of elements besides hydrogen and helium. However, the very first generation of star syste

Mariulka [41]

Answer:

Answer

Explanation:

The very first generation of star systems in the universe probably consisted only of hydrogen and helium. This means that,

There were no comets or asteroids in these first-generation star systems. As the comets and asteroids were the only soruce of elements other than Hydrogen and Heliums ( As starts mainly consists of helium and hydrogen).

3 0

3 years ago

In a crude model of a rotating diatomic molecule of chlorine (Cl2), the two Cl atoms are 2.00 ✕ 10-10 m apart and rotate about t

Troyanec [42]

Answer:

$1.03723\times 10^{-20}\ J$

Explanation:

r = Radius of atom = $1\times 10^{-10}\ m$

m = Mass of chlorine atom = $5.88\times 10^{-26}\ kg$

$\omega$ = Angular speed = $4.2\times 10^{12}\ rad/s$

The moment of inertia of the system is given by

$I=2mr^2\\\Rightarrow I=2\times 5.88\times 10^{-26}\times (10^{-10})^2\\\Rightarrow I=1.176\times 10^{-45}\ kgm^2$

Kinetic energy is given by

$K=\dfrac{1}{2}I\omega^2\\\Rightarrow K=\dfrac{1}{2}\times 1.176\times 10^{-45}\times (4.2\times 10^{12})^2\\\Rightarrow K=1.03723\times 10^{-20}\ J$

The rotational kinetic energy of one molecule of the atom is $1.03723\times 10^{-20}\ J$

7 0

3 years ago

A truck is hauling a 300-kg log out of a ditch using a winch attached to the back of the truck. Knowing the winch applies a cons

Pie

Answer:

0.128 s

Explanation:

We have to start by calculating the net force acting on the log. We have two forces:

- The constant pulling force, forward, of F = 2500 N

- The frictional force, backward

The frictional force is given by

$F_f = \mu mg$

where

$\mu=0.45$ is the coefficient of friction

m = 300 kg is the mass of the log

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting,

$F_f = (0.45)(300)(9.8)=1323 N$

So the net force acting on the log is

$F=2500 - 1323=1177 N$

Now, we can find the acceleration of the log by using Newton's second law

$F=ma$

where a is the acceleration. Re-arranging for a,

$a=\frac{F}{m}=\frac{1177}{300}=3.92 m/s^2$

And finally we can find the time it takes for the log to reach a speed of

v = 0.5 m/s

by using the suvat equation:

$v=u+at$

where u = 0 is the initial speed and t the time. Solving for t,

$t=\frac{v}{a}=\frac{0.5}{3.92}=0.128 s$

5 0

4 years ago

A transverse, wave travelling on a chord is represented by D=0.22sin (5.6x+34t) where D and x are inmeters and t is in seconds.

ArbitrLikvidat [17]

Answer:

a) λ = 1.12 m

b) f = 5.41 Hz

c) v = 154.54 m/s

d) A = 0.22m

$v_D_{max}=7.48\frac{m}{s}\\\\v_D_{min}=-7.48\frac{m}{s}\\\\$

Explanation:

You have the following equation for a wave traveling on a cord:

$D=0.22sin(5.6x+34t)$ (1)

The general expression for a wave is given by:

$D=Asin(kx-\omega t)$ (2)

By comparing the equation (1) and (2) you have:

A: amplitude of the wave = 0.22m

k: wave number = 5.6 m^-1

w: angular velocity = 34 rad/s

a) The wavelength is given by substitution in the following expression:

$\lambda=\frac{2\pi}{k}=\frac{2\pi}{5.6m^{-1}}=1.12m$

b) The frequency is:

$f=\frac{\omega}{2\pi}=\frac{34s^{-1}}{2\pi}=5.41Hz$

c) The velocity of the wave is:

$v=\frac{\omega}{k}=\frac{34s^{-1}}{0.22m^{-1}}=154.54\frac{m}{s}$

d) The amplitude is 0.22m

e) To calculate the maximum and minimum speed of the particles you obtain the derivative of the equation of the wave, in time:

$v_D=\frac{dD}{dt}=(0.22)(34)cos(5.6x+34t)\\\\v_D=7.48cos(5.6x+34t)$

cos function has a minimum value -1 and maximum +1. Then, you obtain for maximum and minimum velocity:

$v_D_{max}=7.48\frac{m}{s}\\\\v_D_{min}=-7.48\frac{m}{s}\\\\$

6 0

3 years ago