The manufacturer of a new laser claims that with only two standard aa batteries (together: 3.0 v, 20 kj energy storage) that the product life is 25 hours. what is the approximate resistance in the laser’s electrical circuitry?

The resistance of the leser = 40.5 Ω

Explanation:

Energy stored in a battery (E) = V²t/R Equation 1

Where V= potential difference of the batteries, t = time, and R = resistance of the batteries.

Making R the subject of equation 1 above,

R = V²t/E substituting this values into equation 2

Where V= 3.0V,  E = 20 kJ = 20 × 1000 = 20000 J, t = 25 hours = 25× 3600 = 90000 s

Substituting this values into equation 2,

R = 3²(90000)/20000

R = 40. 5 ohms (Ω)

The resistance of the leser = 40.5 Ω

the answer is the index of refraction of the second medium is higher. i just took the test.