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Andrej [43]
3 years ago
11

Suppose there is a 3Mbps uplink and 10Mbps downlink between a geostationary satellite and the base station on earth. If the prop

agation speed is the speed of light (3 * 10^8 m/sec), the packet size is 20Mb, and the distance from the satellite to earth is 36,000 km, what is the uplink propagation delay of the link in milliseconds
Physics
1 answer:
Angelina_Jolie [31]3 years ago
6 0

Answer:

120\ \text{ms}

Explanation:

Distance between the satellite and Earth = 36000\ \text{km}

Speed of light = 3\times 10^8\ \text{m/s}

Propagation delay is given by distance by the speed of light

\dfrac{36000\times 10^3}{3\times 10^8}

=0.12\ \text{s}\times 10^3

=120\ \text{ms}

The uplink propagation delay of the link is 120\ \text{ms}.

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Un pez llamado PARGO ROJO vive a grandes profundidades. Si se pesca, al salir a la superficie puede tomar el aspecto de la foto
zalisa [80]

Answer:

Hay diversas leyes que podemos usar acá.

Acá sabemos que la vejiga aumenta su tamaño al reducir la presión, esto tiene sentido, pues al haber menos presión, hay menos fuerza que comprime la vejiga, lo que le permite aumentar su volumen.

Acá tenemos una relación inversa de la forma: V = K/P

Una relación inversa donde la presión esta en el denominador y K es un termino que no depende ni del volumen ni de la presión.

Entonces, a medida que aumenta P, el denominador aumenta, por lo que el valor del volumen decrece.

Un ejemplo de una ecuación similar es la del gas ideal, por ejemplo, para un gas ideal dentro de un globo de volumen V para una dada presión P:

V = nRT/P

donde n es el numero de moles, R es la constante termodinámica y T es la temperatura, acá podemos ver que esta ecuación tiene la misma forma fundamental que la escrita arriba.

7 0
3 years ago
Since rods are about 1000 times more sensitive than cones (at 470 nm), they should be able to detect smaller values of the elect
8090 [49]

Answer:

  a. Rods are about 1000 times more sensitive than cones.

Explanation:

The answer choice shown is a direct copy of the first line of the problem statement. It doesn't appear to be any more complicated than that. (It's a reading comprehension question.)

3 0
3 years ago
Two large, parallel, nonconducting sheets of positive charge face each other. What is at points (a) to the left of the sheets, (
alexira [117]

Answer:

a)The electric Field will be zero at the point between the sheets

b)E_1=\dfrac{\sigma}{\epsilon_0}

c)E_2=\dfrac{\sigma}{\epsilon_0}

Explanation:

Let \sigma be the surface charge density of the of the non conducting parallel sheet.Let consider a Gaussian surface in the form of of cylinder such that its cross-sectional is A . Then there will be flux only due to cross sectional area as the curved sectional is perpendicular to the the electric field  so the Electric Flux due to it is zero.

Now using Gauss law we have, E be the electric Field at the distance r from the sheet then

E\times 2A=\dfrac{\sigma A}{\epsilon_0}\\E=\dfrac{\sigma}{2\epsilon_0}

The Field will be away from the sheet and perpendicular to it.

a) The Electric Field between them

E_1=\dfrac{\sigma}{2\epsilon_0}-\dfrac{\sigma}{2\epsilon_0}\\=0

b)The Electric Field to the right of the sheets

E_1=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}\\=\dfrac{\sigma}{\epsilon_0}

c)The Electric Field to the left of the sheets

E_2=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}\\=\dfrac{\sigma}{\epsilon_0}

3 0
3 years ago
Please help me do this problem
Digiron [165]

Answer:

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for 4 time-steps (2 to 6) the velocity is 6 per time step, that makes 24 distance units in these 4 time steps. it's the same the area underneath the graph.

there is also the vertical line from 0 to 2. we can calculate that distance like the area of a triangle with 2*6 / 2 = 6

the total distance from 0 to D is therefore 30

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