. A lightbulb with a resistance of 2.9 ohms is operated using a 1.5-volt battery. At what rate is
electrical energy transformed in the lightbulb?

Explanation:

Given that,

A light bulb has a resistance of 2.9ohms

R = 2.9 ohms

And a battery of 1.5V is applied

V = 1.5 V

We want to find the rate of energy transformed

First we need to know what rate of energy is

Rate of energy implies that we want to find power. Power is the rate at which work is done

P = Workdone / time

Then,

In electronic, the power dissipated by a resistor is given as

P = V² / R

P = 1.5² / 2.9

P = 0.7759 W

P ≈ 0.776 W

So, the rate at which electrical energy transformed in the lightbulb is 0.776 Watts

The rate of electrical energy transformed in the light bulb is 0.78 Joules per seconds.

Explanation:

Electrical energy is given by,

       E = IVt

The rate of electrical energy transformed in the light bulb can be determined by;

             = IV

From Ohms law which states that the amount of current passing through a metallic conductor, e.g wire, is directly proportional to the potential difference across its ends, provided that the temperature is constant.

  i.e       V = IR

where: V is the potential difference, I is the current and R is the resistance of the conductor

So that: R = 2.9 Ohms and V = 1.5 Volt, find I.

⇒              I =  

                  =

                  = 0.5172

                I = 0.52 A

Then;

         = IV

            = 0.52 × 1.5

            = 0.78 J/s

The rate of electrical energy transformed in the light bulb is 0.78 Joules per seconds.

32: from largest to smallest, this hierarchy includes eons, eras, periods, epochs, and ages.

33: answer is c

i really hope this answer you out! it makes my day people like you and giving back to the community that has me through school! if you could do me a favor, if this you and this is the very best answer and you understand that all of my answers are legit and top notch. mark as brainliest! and have a awesome day!