Answer: The distance of Observer A from the radio antenna is what fraction of the distance of Observer B from the radio antenna?
It is 1/4.
Step-by-step explanation:
We know that the intensity of electromagnetic waves decreases with the radius squared, this means that we can write a simple relation as:
Intensity(r) = A/r^2
Observer A measures 16 the intensity of observer B.
if Ia is the intensity that observer A measures and Ib is the intensity that observer B measures, we have that:
Ia = 16Ib
A/(ra)^2 = 16*A/(rb)^2
1/(ra)^2 = 16/(rb)^2
rb^2 = 16*ra^2
and we know that 16 = 4*4 = 4^2
rb^2 = (4*ra)^2
then rb = 4*ra
this means that the distance between observer B and the antenna is equal to 4 times the distance between observer A and the antenna.
The fraction is ra = rb/4
The distance of
Observer A from the radio antenna is what fraction of the distance of Observer B from the radio antenna?
It is 1/4.
Answer:
127
Step-by-step explanation:
Answer:
unit vector = -0.891i + 0.454j
Step-by-step explanation:
Given that;
magnitude of unit vector =
angle of 153° degrees from the x axis, with a positive y component
{ second quadrant }
so, x and y components of the vector are;
x component = 1 × cos(153°) = - 0.891
y component = 1 × sin(153°) = 0.454
⇒ -0.891i + 0.454j
Therefore, unit vector = -0.891i + 0.454j
-10 because on a number line once you get to the negatives the numbers go backwards.