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Peter drove at 45 kmhr for 3 hours and at 30 kmhr for the rest of the journey of 195 kms. How long did he drive at 30 kmhr?
2 hours
What is average speed If a man travels 20 km on foot at 5 kmhr and another man10 km by bus at 20 kmhr?
The average speed for both men (together) is 6.66... (repeating) km per hour.
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
An airplane flew for 6 hours against a tail wind of 25 kmhr. The return trip with the same wind took 5 hours. Find the speed of the airplane in still air?
275 km/hr
If a boat is traveling east at 20 kmhr and the river it is on flows west at 5 kmhr what is the velocity of the boat?
20 km/hr - 5 km/hr = 15 km/hr
170.Walking at 4 kmhr a man reaches his office 5 minutes late. If he walks at 5 kmhr he will be 2.5 minutes early. What is the distance of his office from his house?
It is 2.5 km.
When was KMHR created?
KMHR was created on 1981-10-26.
How do you transfer 30 seconds into a millisecond?
1000 msec = 1 sec 30,000 msec = 30 sec
Peter drove at 45 kmhr for 3 hours and at 30 kmhr for the rest of the journey of 195 kms. How long did he drive at 30 kmhr?
2 hours
What is average speed If a man travels 20 km on foot at 5 kmhr and another man10 km by bus at 20 kmhr?
The average speed for both men (together) is 6.66... (repeating) km per hour.
How many msec in sec?
1000
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
How fast is 275 kmhr?
170.88 MPH
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%
