Ca +2
At -1
In +3
Sr +2
Ra +2
Fr +1
Ba +2
As -3
Answer:

Explanation:
Although the context is not clear, let's look at the oxidation and reduction processes that will take place in a Fe/Sn system.
The problem states that anode is a bar of thin. Anode is where the process of oxidation takes place. According to the abbreviation 'OILRIG', oxidation is loss, reduction is gain. Since oxidation occurs at anode, this is where loss of electrons takes place. That said, tin loses electrons to become tin cation:

Similarly, iron is cathode. Cathode is where reduction takes place. Reduction is gain of electrons, this means iron cations gain electrons and produce iron metal:

The net equation is then:

However, this is not the case, as this is not a spontaneous reaction, as iron metal is more reactive than tin metal, and this is how the coating takes place. This implies that actually anode is iron and cathode is tin:
Actual anode half-equation:

Actual cathode half-equation:

Actual net reaction:

Friction of the air, and the surface it is on
Answer:
0.6 Ω
Explanation:
From the question given above, the following data were obtained:
Voltage (V) = 12 V
Current (I) = 20 A
Resistance (R) =?
From Ohm's law,
V = IR
Where:
V => is the voltage
I => is the current
R => R is the resistance
With the above formula, we can obtain the resistance as follow:
Voltage (V) = 12 V
Current (I) = 20 A
Resistance (R) =?
V = IR
12 = 20 × R
Divide both side by 20
R = 12 / 20
R = 0.6 Ω
Thus the resistance is 0.6 Ω
I think that the answer is (D)... I hope this helped