Unpolarized light shines pon a photocell for one hour. then, two polarizers, whose transmission axies are offset by 60, are placed in front of the photocell such that the light must first pass through the polarizers before reaching the cell.
How much time is required for the photocell to receive the same total energy as it received in one hour’s time without the polarizers present?

  t = 1.10 h

Explanation:

For this exercise we must calculate the intensity that reaches the cell with the polarizers, for this we use Malus's law

          I = I₀ cos² θ

Let's calculate

         I = I₀ cos² 60

         I = I₀ 0.907

This is the intensity that reaches the cell

To find time, use an inverse proportion rule or rule of three, the intensity Io in

        t = I₀ / (0.907 I₀) 1 h

        t = 1.10 h

wvbbvvv

hello, i am papaguy.

i have found an answer to your question but use it as information.

information: because weak acids/bases have low concentration of hydrogen ions, the addition of water has a large impact on the ph. 2. a difference in 1 unit is a 10x difference in concentration for example; a liquid with ph of 3 is 10x more acidic than a liquid with a ph of 4. therefore, a liquid with a ph of 3 is stronger.

best of luck!