For the same signal emitted by a radio antenna,
Observer A measures its intensity to be 16 times the
intensity measured by Observer B. The distance of
Observer A from the radio antenna is what fraction
of the distance of Observer B from the radio
antenna?

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.

i think the answer is .36

step-by-step explanation:

what i did was only 2.88 divided by 8

answer: add the !

step-by-step explanation: then we will answer