What are the energy transformations involved in Fax/Modem and Radio/Television?

AI Thread Summary
The discussion focuses on the energy transformations in fax/modem and radio/television systems. For fax and modem, the process involves mechanical energy from moving components, which converts to electrical energy, then to signal energy, and back to mechanical. In radio and television, energy transformations typically include electromagnetic energy converting to electrical energy and then to sound or visual output. The complexity of these transformations can vary based on the technology used. Understanding these basic transformations is essential for grasping how these devices operate.
CustomPker
Messages
4
Reaction score
0

Homework Statement



describe the energy transformations required in the following

Fax/Modem
Radio And Television

thanks
 
Physics news on Phys.org
not sure how complex to go(what grade are you in). basically
electrical to mechanical
electromagnetic-electrical-sound there is probably more but you get the jist.
 
currently year 11,
 
i just don't understand the energy transformations of the fax/modem
 
the fax is first off mechanical (moving cmpts) then to electrical then depending on the fax some sort of signal energy then back to mechanical. the modem is the same as the last part of the fax, I am assuming you haven't done any signal processing in yr 11. That is all your teacher would want.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top