• Congratulations to the Class of 2024 on your results!
    Let us know how you went here
    Got a question about your uni preferences? Ask us here

Question... bit tricky (1 Viewer)

currysauce

Actuary in the making
Joined
Aug 31, 2004
Messages
576
Location
Sydney
Gender
Male
HSC
2005
The current of an a.c source reverses its polarity every 0.010 micro seconds. THe wavelength of the radio waves produced is what?
 
Last edited:

rukie

5th November, Here I Come
Joined
Oct 13, 2004
Messages
8
Gender
Male
HSC
2005
Im not entirely sure, but i came across a similar question so ill use the same method

Well, it changes polarity every 0.01 microseconds, so for a full cycle, it must change polarity twice, so double that you get 0.02, thers your period, frequency = 1/p = 5x10^7 Hz (changing microsecs to seconds)

Using wavelength = c/frequency u get 6 m
 
Last edited:

Jago

el oh el donkaments
Joined
Feb 21, 2004
Messages
3,691
Gender
Male
HSC
2005
wow...that's clever
 

Xayma

Lacking creativity
Joined
Sep 6, 2003
Messages
5,953
Gender
Undisclosed
HSC
N/A
Well you don't see the axioms for maths either. You should know it by now.
 

helper

Active Member
Joined
Oct 8, 2003
Messages
1,183
Gender
Male
HSC
N/A
powerhouse said:
I never saw the Period = 1 / frequency formula in the syllabus.
Its not a formula given but is assumed knowledge from the year 11 section on the world communicates.
 

jarro_2783

Member
Joined
Dec 1, 2004
Messages
63
Location
Australia
Gender
Male
HSC
2005
even if you don't know that formula, think about it this way.

frequency is cycles per second. So you know that one cycle is 0.02 micro seconds.
ie: f = 1 (cycle) / 0.02 (microseconds), convert to seconds and just do that division.

You can use that for any measurement per another measurement calculation like metres per second, watts per metre or whatever. If you've got one measurement per another measurement and you know a value for both, just put it in the calculator and that is your answer for the first measurement per unit of the second one.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top