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2 years ago

Why are the wires twisted around each other in twisted pair cable?

1 answer:

5 0

The wires are twisted around each other in a twisted pair cable for the purpose of blocking off any external electromagnetic interference.

A twisted-pair cable is a form of cable system used for telecom services as well as most current wired networks. Twisted pairs are composed of two insulated copper wires that are been twisted together. The circuit is formed by a twisted pair of wires that may carry data. These pairs are twisted to prevent interference or noise caused by neighboring pairs.

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Describe four energy changes that happen in the process.

musickatia [10]

Driving a motor........

chemical energy is converted into kinetic energy.

Falling off of cliff

.........gravitational potential energy is converted into kinetic energy.

Hydroelectric energy generation

.......gravitational potential energy is converted into kinetic energy (i.e. driving a generator), which is then converted into electrical energy.

Nuclear power generation

.........mass is converted into energy, which then drives a steam turbine, which is then converted into electrical energy.

7 0

2 years ago

An 1700 kg car is moving to the right at a constant speed of 1.50 m/s. (a) what is the net force on the car

cluponka [151]

I am sorry if I am wrong but, the net force would be zero. 0

8 0

3 years ago

¿Por que una una tapa pequeña de agua purificada se mantiene quieta por más tiempo sobre una rampa, en comparación al tiempo red

shusha [124]

Answer:

Esto sucede principalmente por el movimiento rotacional de las ruedas del carrito.

Como sabemos, existen dos tipos de coeficientes de fricción, el estacionario y el cinético.

En principio, la tapa se mantendrá quieta por que no tendrá la suficiente fuerza como para vencer al coeficiente de fricción estático, por lo que no podrá moverse hasta que esta reciba un pequeño impulso, como puede ser una pequeña corriente de aire. (asumo que la superficie curva de la tapita es la que esta en contacto con la rampa)

(Voy a simplificar el movimiento rotacional, pero creo que es suficiente para explicar la situación)

Ahora, en el carrito, las ruedas están conectadas a un eje.

La fuerza gravitatoria podemos pensarla que esta arriba de este eje, mas o menos en el centro del carrito, y esta fuerza intentara que el carrito baje de la rampa (lo mismo pasa para la tapita), pero a diferencia de la tapita, esta fuerza es aplicada a una distancia dada del eje (o del centro de las rueditas) lo que podemos pensar que funciona como una palanca, amplificando así la fuerza, por lo que esta vez el rozamiento estático podrá detener la superficie de la rueda que esta en contacto con el suelo, pero cuanto el centro de masa se mueva un poco, el punto de contacto entre la rueda y el suelo se moverá, de esta manera aparece lo que llamamos movimiento rotacional (si no hubiera fricción, las ruedas simplemente deslizarían sobre la rampa) que le permite al carrito ignorar en parte al coeficiente de fricción estático.

Otro factor importante, es que las tapitas suelen tener una superficie rugosa y un centro de masa desfazado, lo que no las hace buenas rotadoras.

5 0

2 years ago

Question 2

Ksivusya [100]

5.77 × $10^1^4$ Hz is the green photon's frequency .

The distance between similar points (adjacent crests) in adjacent cycles of a waveform signal that is propagated in space is known as the wavelength. A wave's wavelength is often measured in meters (m), centimeters (cm), or millimeters (mm) (mm). The relationship between frequency and wavelength is inverse.

<h3>Given:</h3>

Wavelength of green light = 520 nm

f = c / λ

where, f = Frequency

c = Speed of light = 3 × $10^8$ m/s

λ = Wavelength of light

∴ f = c / λ

f = $\frac{3*10^8}{520 * 10^-^9}$

= 5.77 × $10^1^4$ Hz

Therefore, 5.77 × $10^1^4$ Hz is the green photon's frequency .

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4 0

1 year ago

I need answers and solvings to these questions

den301095 [7]

1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)

2) The angular acceleration is C) $15.7 rad/s^2$

3) The frequency of the oscillation is C) 1.6 Hz

4) The period of vibration is B) 0.6 s

5) The diameter of the nozzle is A) 5.0 mm

6) The force that must be applied is B) 266.7 N

Explanation:

The period of a simple pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that the period of the pendulum depends only on its length and on the acceleration of gravity, while there is no dependence on the mass of the pendulum or on the amplitude of oscillation. Therefore, the correct option is

B) III. only (the length of the pendulum)

The angular acceleration of the rotating disc is given by the equation

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f$ is the final angular velocity

$\omega_i$ is the initial angular velocity

t is the time elapsed

For the compact disc in this problem we have:

$\omega_i = 0$ (since it starts from rest)

$\omega_f = 300 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=31.4 rad/s$ is the final angular velocity

t = 2 s

Substituting, we find

$\alpha = \frac{31.4-0}{2}=15.7 rad/s^2$

For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation

$a=-\omega^2 x$

where

a is the acceleration

x is the displacement

$\omega$ is the angular frequency of the system

For the oscillator in this problem, we have the following relationship

$a=-100 x$

which implies that

$\omega^2 = 100$

And so

$\omega = \sqrt{100}=10 rad/s$

Also, the angular frequency is related to the frequency $f$ by

$f=\frac{\omega}{2\pi}$

Therefore, the frequency of this simple harmonic oscillator is

$f=\frac{10}{2\pi}=1.6 Hz$

When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write

$mg=kx$

where

m is the mass

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

k is the spring constant

x = 8.0 cm = 0.08 m is the stretching of the spring

We can re-arrange the equation as

$\frac{k}{m}=\frac{g}{x}=\frac{9.8}{0.08}=122.5$

The angular frequency of the spring is given by

$\omega=\sqrt{\frac{k}{m}}=\sqrt{122.5}=11.1 Hz$

And therefore, its period is

$T=\frac{2\pi}{\omega}=\frac{2\pi}{11.1}=0.6 s$

According to the equation of continuity, the volume flow rate must remain constant, so we can write

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the hose, with $r_1 = 5 mm$ being the radius of the hose

$v_1 = 4 m/s$ is the speed of the petrol in the hose

$A_2 = \pi r_2^2$ is the cross-sectional area of the nozzle, with $r_2$ being the radius of the nozzle

$v_2 = 16 m/s$ is the speed in the nozzle

Solving for $r_2$ , we find the radius of the nozzle:

$\pi r_1^2 v_1 = \pi r_2^2 v_2\\r_2 = r_1 \sqrt{\frac{v_1}{v_2}}=(5)\sqrt{\frac{4}{16}}=2.5 mm$

So, the diameter of the nozzle will be

$d_2 = 2r_2 = 2(2.5)=5.0 mm$

According to the Pascal principle, the pressure on the two pistons is the same, so we can write

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

where

$F_1$ is the force that must be applied to the small piston

$A_1 = \pi r_1^2$ is the area of the first piston, with $r_1= 2 cm$ being its radius

$F_2 = mg = (1500 kg)(9.8 m/s^2)=14700 N$ is the force applied on the bigger piston (the weight of the car)

$A_2 = \pi r_2^2$ is the area of the bigger piston, with $r_2= 15 cm$ being its radius

Solving for $F_1$ , we find

$F_1 = \frac{F_2A_1}{A_2}=\frac{F_2 \pi r_1^2}{\pi r_2^2}=\frac{(14700)(2)^2}{(15)^2}=261 N$

So, the closest answer is B) 266.7 N.

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5 0

3 years ago