Cable transmission of signals
HS Varun asked:
How is it possible to transmit a no. of channels through the same cable ?
Is it based on the concept of Bandwidth ?
Answer:
In a cable TV system, each channel is assigned a range of frequency (6-8MHz range)
Categories: Communication Systems Tags: answer, ask Physics, communication, concept, message, physics, range, signal, system
Radiations sensed by eyes
Arun Asked:
“What form of radiation can your eyes detect? sir, please answer this question”
Answer:
Normal Human Eyes can sense the visible spectrum only.
“The visible spectrum is the portion of the electromagnetic spectrum that is visibleto (can be detected by) the human eye. Electromagnetic radiation in this range ofwavelengths is called visible light or simply light. A typical human eye will respond to wavelengths from about 390 to 750 nm.[1] In terms of frequency, this corresponds to a band in the vicinity of 400–790 THz.” ———– Wikipedia
Categories: Uncategorized Tags: ask Physics, form, human eye, light, Normal Human Eyes, physics, radiation, range, spectrum
Kinematics Numerical
- A particle travels 20 m in 7th second and 24m in 9th sec. find initial velocity?
- in a projectile motion , a body thrown from the ground ,at what angle both the vertical height and range will be equal?
- The velocity v(cm/s) of a particle is given in terms of time t(in seconds) by the equation v = at+b/t+c. Dimension of a,b, and c are ?
Categories: Ask Physics Tags: ask, ask Physics, Auto, body, doubt, Draft, ground, height, initial velocity, kinematics, numerical, Online, particle, physics, project, projectile, projectile motion, range, sec, second, solved, solved paper, solved problems, solved questions, time t, velocity
A Problem from Projectile motion
“A ball is projected vertically upwards with an initial velocity of 28m/s and is observed to pass a point 30m above the projection point,at what 2times does the ball pass these points?”
(Vimal Raj Answered)
ans : 1.356 sec and 1.444 sec .
sol:
from v=u+at we get t=2.8 . (time taken to reach the max height )
fm v2-u2=2aS we get S=39.2 m (max height reached)
observer at 30 m .
so , in btw distance btw observer and ball (at the top) =9.2 m
at top ball is at rest and time taken to reach the observer can be calculaed from S=ut + 1/2 at2 ,we get t=1.356s .
and next time can be calculated by 2.8 – 1.356 =1.444 sec .
(The answer has not been scrutinized. If any error, teachers and visitors can post as comment)
Categories: KINEMATICS, Mechanics, Project Tags: kinematics, numerical, problem, projectile motion, range