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Lelu [443]
3 years ago
13

- How long does it take a packet of length 1,000 bytes to propagate over a link of distance 2,500 km, propagation speed 2.5 *10^

8 m/s, and transmission rate 2 Mbps? More generally, how long does it take a packet of length L to propagate over a link of distance d, propagation speed s, and transmission rate R bps? Does this delay depend on packet length? Does this delay depend on transmission rate?
Physics
1 answer:
Sunny_sXe [5.5K]3 years ago
3 0

Answer:

It will take 0.01 s or 10 ms

Solution:

As per the question:

Length of the packet, L = 1,000 bytes = 1000\times 8 = 8000 bits

Distance, d = 2500 km = 2.5\times 10^{6}\ m

Speed of propagation, s = 2.5\times 10^{8}\ m/s

Transmission rate, R = 2 Mbps

Now,

Propagation time, t can be calculated as:

t = \frac{d}{s} = \frac{2.5\times 10^{6}}{2.5\times 10^{8}} = 0.01\ s

t = 10 ms

  • In general, propagation time, t is given by:

       t = \frac{link\ distance}{Propagation\ speed}

  • No, this delay is independent of the length of the packet.
  • No, this delay is independent of the rate of transmission.

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Some of the highest tides in the world occur in the Bay of Fundy on the Atlantic Coast of Canada. At Hopewell Cape the water dep
Nady [450]

Answer:

(a) 1.939 m/h

(b) 0.926 m/h

(c) -0.315 m/h

(d) -1.21 m/h

Explanation:

Here, we have the water depth given by the function of time;

D(t) = 7 + 5·cos[0.503(t-6.75)]

Therefore, to find the velocity of the depth displacement with time, we differentiate the given expression with respect to time as follows;

D'(t) = \frac{d(7 + 5\cdot cos[0.503(t-6.75)])}{dt}

= 5×(-sin(0.503(t-6.75))×0.503

= -2.515×(-sin(0.503(t-6.75))

= -2.515×(-sin(0.503×t-3.395))

Therefore we have;

(a) at 5:00 AM = 5 -  0:00 = 5

D'(5) =  -2.515×(-sin(0.503×5-3.395)) = 1.939 m/h

(b) at 6:00 AM = 6 -  0:00 = 6

D'(5) =  -2.515×(-sin(0.503×6-3.395)) = 0.926 m/h

(c) at 7:00 AM = 7 -  0:00 = 7

D'(5) =  -2.515×(-sin(0.503×7-3.395)) = -0.315 m/h

(d) at Noon 12:00 PM = 12 -  0:00 = 12

D'(5) =  -2.515×(-sin(0.503×12-3.395)) = -1.21 m/h.

4 0
2 years ago
An electron accelerated from rest through a voltage of 780 v enters a region of constant magnetic field. part a part complete if
maxonik [38]
The electron is accelerated through a potential difference of \Delta V=780 V, so the kinetic energy gained by the electron is equal to its variation of electrical potential energy:
\frac{1}{2}mv^2 =  e \Delta V
where
m is the electron mass
v is the final speed of the electron
e is the electron charge
\Delta V is the potential difference

Re-arranging this equation, we can find the speed of the electron before entering the magnetic field:
v= \sqrt{ \frac{2 e \Delta V}{m} } = \sqrt{ \frac{2(1.6 \cdot 10^{-19}C)(780 V)}{9.1 \cdot 10^{-31} kg} }=1.66 \cdot 10^7 m/s


Now the electron enters the magnetic field. The Lorentz force provides the centripetal force that keeps the electron in circular orbit:
evB=m \frac{v^2}{r}
where B is the intensity of the magnetic field and r is the orbital radius. Since the radius is r=25 cm=0.25 m, we can re-arrange this equation to find B:
B= \frac{mv}{er}= \frac{(9.1 \cdot 10^{-31}kg)(1.66 \cdot 10^7 m/s)}{(1.6 \cdot 10^{-19}C)(0.25 m)} =3.8 \cdot 10^{-4} T
3 0
3 years ago
A 22kg Accelerates at a rate of 2.3 m/s. What is the magnitude of the net force acting on the bike?
Tcecarenko [31]

magnitude of the net force = mass x acceleraton

                                             = 22 x 2.3

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7 0
3 years ago
a concave mirror has a focal length of 18 cm. where will an image form if an object is placed 58 cm from the mirror
MAXImum [283]

Answer:

here

Explanation:

8 0
3 years ago
A 26 kg body is moving through space in the positive direction of an x axis with a speed of 350 m/s when, due to an internal exp
Gnom [1K]

Answer:

a.) 1567.2 m/s

b.) 149.4 m/s

Explanation:

Given that a 26 kg body is moving through space in the positive direction of an x axis with a speed of 350 m/s when, due to an internal explosion, it breaks into three parts. One part, with a mass of 7.8 kg, moves away from the point of explosion with a speed of 180 m/s in the positive y direction. A second part, with a mass of 8.8 kg, moves in the negative x direction with a speed of 640 m/s.

The x-component of the third part can be calculated by assuming that it moves in a positive x axis.

The third mass = 26 - ( 7.8 + 8.8)

The third mass = 26 - 16.6

The third mass = 9.4kg

since momentum is conserved, the momentum before explosion will be equal to sum of the momentum after explosion

26 x 350 = -8.8 x 640 + 9.4V

9100 = -5632 + 9.4V

9.4V = 9100 + 5632

9.4V = 14732

V = 14732/9.4

V = 1567.2 m/s

(b) y-component of the velocity of the third part will be

7.8 x 180 = 9.4 V

1404 = 9.4V

V = 1404/9.4

V = 149.4 m/s

7 0
2 years ago
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