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How do you transfer 30 seconds into a millisecond?
1000 msec = 1 sec 30,000 msec = 30 sec
What is delay sensitive and bandwidth sensitive operations?
delay sensitive is a time delay parameter which is oftenly used in multimedia applications. The units of this parameter is sec. It usually ranges from 100 msec to 5 sec.
How do you calculate seconds to msec?
Each second has 1,000 msec in it. Multiply the number of seconds by 1,000 and you have the number of milliseconds in the same length of time.
What is the Standard abbreviation for milliseconds?
ms or msec
Write 3.5 seconds as milliseconds?
3500 msec
What is the mass of a runner moving at the speed of 2.5 msec with a momentum of 200 kg m per sec?
Momentum is defined as mass * velocity, M=m*v. Momentum: 200kg*m/sec velocity: 2.5m/sec mass: (200kg*m/sec) / (2.5m/sec) = 80kg If remembering equations like these is difficult, then something useful for you would be to understand the concept of dimensional analysis. That is a fancy word for looking at the units of the given quantities. If you know that mass is kilograms, then you can figure out how to manipulate the units m/sec and kg*m/sec in order to come out with a final unit of kg. By dividing kg*m/sec with m/sec you end up getting (kg*m*sec) / (m*sec), leaving just kg left which is the unit you were looking for. Therefore you should divide 200 by 2.5 for a final answer of 80kg.
Frames of 1000 bits are sent over a 1-Mbps channel using a geostationary satellite whose propagation time from the earth is 270 msec?
Let t=0 denote the start of transmission. At t=1 msec, the first frame has been fully transmitted. At t=271 msec, the first frame has fully arrived. At t=272 msec, the frame acknowledging the first one has been fully sent. At t=542 msec, the acknowledementbearing frame has fully arrived. Thus, the cycle is 542 msec. A total of k frames are sent in 542 msec, for an efficiency of k/542. Hence a) k=1, efficiency=1/542=0,18% b) k=7, efficiency=7/542=1,29% c) k=4, efficiency=4/542=0,74%
What is a round trip delay in satellite communication?
250 msec
How many sec are in 5 min?
300 sec
Consider an error free 64-Kbps satellite channel used to send 512 byte data frames in one direction with very short acknowledgements coming back the other way?
The transmission starts at t = 0. At t = 4096/64000 sec = 64 msec, the last bit is sent. At t = 334 msec, the last bit arrives at the satellite and the very short ACK is sent. At t = 604 msec(270 + 334), the ACK arrives at the earth. The data rate here is 4096 bits in 604 msec or about 6781 bps (4096/604msec) (window size is 1). With a window size of 7 frames, transmission time is 448 msec(512*8*7/64000) for the full window, at which time the sender has to stop. At 604 msec, the first ACK arrives and the cycle can start again. Here we have 7 × 4096 = 28,672 bits in 604 msec. The data rate is 47,470.2 bps (28,672/604msec) (window size is 7). Continuous transmission can only occur if the transmitter is still sending when the first ACK gets back at t = 604 msec. In other words, if the window size is greater than 604 msecworth of transmission, it can run at full speed. For a window size of 10 or greater, this condition is met, so for any window size of 10 or greater (e.g., 15 or 127), the data rate is 64 kbps.
What is the website of maharashtra state board of education?
www.msbshse.ac.in ORwww.mah.nic.in/msec
How long is the latent period when a muscle twitch?
around 4 msec
