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oee [108]
3 years ago
14

When equilibrium is reached in an electrochemical cell, the voltage reaches its maximum value. 1. false 2. true?

Physics
2 answers:
tigry1 [53]3 years ago
7 0

False

Voltage in an electrochemical cell is in indication of equilibrium, higher will be the non-equilibrium, higher will be the voltage, or we can say at equilibrium voltage tends to 0.

Voltage in an electrical cell is the result of flow of electron, which flow due to difference in charge of the cells, higher the charge difference higher will be the voltage, as the equilibrium between the chemical cells established the flow of electron will stop, and the voltage of the cell tend to 0.

WITCHER [35]3 years ago
7 0

Answer: The given statement is false.

Explanation:

Voltage in an electrochemical cell helps in determining that the system is out of equilibrium.

When a redox reaction approaches towards equilibrium then there occurs flow of electrons with the advancement in chemical reaction. This means as the chemical reaction proceeds towards completion then flow of electrons also takes place.

Hence, cell potential decreases until the reaction reaches at equilibrium where, \Delta G = 0.

Also,  \Delta G = -nFE. So, at equilibrium voltage becomes equal to zero and current also stops.

Thus, we can conclude that the statement when equilibrium is reached in an electrochemical cell, the voltage reaches its maximum value is false.

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Assume that the speed of light in a vacuum has the hypothetical value of 18.0 m/s. A car is moving at a constant speed of 14.0 m
denis23 [38]

Answer:

4.245s

Explanation:

Given that,

Hypothetical value of speed of light in a vacuum is 18 m/s

Speed of the car, 14 m/s

Time given is 6.76 s, and we're asked to find the observed time, T

The relationship between the two times can be given as

T = t / √[1 - (v²/c²)]

The missing variable were looking for is t, and we can find it if we rearrange the formula and make t the subject

t = T / √[1 - (v²/c²)]

And now, we substitute the values and insert into the equation

t = 6.76 * √[1 - (14²/18²)]

t = 6.76 * √[1 - (196/324)]

t = 6.76 * √(1 - 0.605)

t = 6.76 * √0.395

t = 6.76 * 0.628

t = 4.245 s

Therefore, the time the driver measures for the trip is 4.245s

8 0
2 years ago
Calculate the electric field at the center of a square
pantera1 [17]

Answer:

E_y=1175510.2\ N.C^{-1}

The Magnitude of electric field is in the upward direction as shown directly towards the charge q_1.

Explanation:

Given:

  • side of a square, a=52.5\ cm
  • charge on one corner of the square, q_1=+45\times 10^{-6}\ C
  • charge on the remaining 3 corners of the square,q_2=q_3=q_4=-27\times 10^{-6}\ C

<u>Distance of the center from each corners</u>=\frac{1}{2} \times diagonals

diagonal=\sqrt{52.5^2+52.5^2}

diagonal=74.25\cm=0.7425\ m

∴Distance of center from corners, b=0.3712\ m

Now, electric field due to charges is given as:

E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}

<u>For charge q_1 we have the field lines emerging out of the charge since it is positively charged:</u>

E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}

  • E_1=2938775.5\ N.C^{-1}

<u>Force by each of the charges at the remaining corners:</u>

E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}

  • E_2=E_3=E_4=1763265.3\ N.C^{-1}

<u> Now, net electric field in the vertical direction:</u>

E_y=E_1-E_4

E_y=1175510.2\ N.C^{-1}

<u>Now, net electric field in the horizontal direction:</u>

E_y=E_2-E_3

E_y=0\ N.C^{-1}

So the Magnitude of electric field is in the upward direction as shown directly towards the charge q_1.

8 0
3 years ago
The type of force measured by grocery store spring scale is
alexandr402 [8]

Answer:

Weight

Explanation:

The spring balance is used to measure weight of an object.

8 0
2 years ago
A 1.5m wire carries a 3 A current when a potential difference of 86 V is applied. What is the resistance of the wire?
Iteru [2.4K]
We know, R = V / I
Here, V = 86 V
I = 3 A

Substitute their values, 
R = 86 / 3
R = 28.67 Ohm

In short, Your Answer would be 28.67 Ohms

Hope this helps!
8 0
2 years ago
Most of the energy released by nuclear fission is in the form of: alpha rays, beta rays, or gamma rays?
Sedbober [7]
I think it is <span>Alpha rays.</span>
3 0
2 years ago
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