The question is looking for "ellipse" and "two" to fill in the blanks.
Answer:
a) Batteries and fuel cells are examples of galvanic cell
b) Ag-cathode and Zn-anode
c) Cell notation: Zn(s)|Zn²⁺(aq) || Ag⁺(aq)|Ag(s)
Explanation:
a) A galvanic cell is an electrochemical cell in which chemical energy is converted to electrical energy. The chemical reaction which drives a galvanic cell is a redox reaction i.e. a reduction-oxidation process.
A typical galvanic cell is composed of two electrodes immersed in a suitable electrolyte and connected via a salt bridge. One of the electrodes serves as a cathode where reduction or gain of electrons takes place. The other half cell functions as an anode where oxidation or loss of electrons occurs. Batteries and fuel cells are examples of galvanic cells.
b) The nature of the electrode that will serve as an anode or cathode depends on the value of the standard reduction potential (E⁰) of that electrode. The electrode with a higher or more positive the value of E⁰ serves as the cathode and the other will function as an anode.
In the given case, the E⁰ values from the standard reduction potential table are:
E⁰(Zn/Zn2+) = -0.763 V
E°(Ag/Ag+)=+0.799 V
Therefore, Ag will be the cathode and Zn will be the anode
c) In the standard cell notation, the anode half cell is written on the left followed by the salt bridge '||' and finally the cathode half cell to the right.
Zn(s)|Zn²⁺(aq) || Ag⁺(aq)|Ag(s)
Answer:
voltage across the resistor
Answer:
The coefficient of static friction between the car and the track
u=0.572
Explanation:
We don't know the mass of the car or any other information so the acceleration is the reason to solve the friction coefficient
∑
As we know

Also the center ward direction forces



But now vt relation with the tangential acceleration

replacing


So magnitude of the force can get by

Get the factor to simplify


Solve to u'


Explanation:
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