Aloop circuit has a resistance of r1 and a current of 2.2 a. the current is reduced to 1.4 a when an additional 3.1 ω resistor is added in series with r1.
what is the value of r1? assume the inter- nal resistance of the source of emf is zero.
answer in units of ω.

Okay so here's the approach I took:

The potential difference in each of the circuits must be the same so if we derive equations for both the potential differences we can set them equal to each other and solve for R1:

In the first circuit

V = 2.2(R1)

In the second we have to find the equivalent resistor, since they are connected in series:

1/R1 + 1/R2 + 1/R3... = Rt

We have R2 so...

1/R1 + 1/3.1 = Rt

1/R1 + 0.323 = Rt

So...

V = 1.4(1/R1 + 0.323)

Set those equal:

2.2R1 = 1.4(1/R1 + 0.323)

2.2R1 = 1.4(1/R1) + 0.4522

Now multiply everything by R1 so we can combine like terms:

2.2R1^2 = 1.4 + 0.4522R1

Isolate to form a quadratic

2.2R1^2 - 0.4522R1 - 1.4 = 0

Solving this quadratic:

R1 = 0.90708 or R1 = -0.701

Since R cannot be negative

R1 = 0.907 ohms

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.

the total distance that martina travels that day is

(distance from home to mom's office) plus (distance from mom's office to school)

plus (distance from school to library) plus (distance from library to friend's house)

plus (distance from friend's house to martina's house).

(we don't know what any of these distance actually is.)