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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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Ivenika [448]
I think its A I hope this help thank you!!
4 0
3 years ago
Newton's second law of motion can also be described as...
Irina-Kira [14]

Answer


B. F=ma



Explanation

The Newton's laws of motion tries to explain the how bodies behave and the energy changes when theys are in motion. For the 3 of them to hold, the bodies must be moving in a straight line and with constant velocity.

The second one states that, "the change of momentum of a moving body is directly proportional to the force producing it and it takes place to the direction of force."

From the choices given, the appropriate answer is B. F=ma

6 0
3 years ago
A vehicle reaches a speed of 7.5 m/s over 15 seconds. What is its acceleration if it
denpristay [2]

acceleration = \frac{velocity}{time}  = \frac{7.5}{15}  = 0.5ms^-^2

So the answer is option b.

8 0
3 years ago
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac
matrenka [14]

Answer:

a) 6.1 m

b) 4.6 s

c) 1.326 m/s

d) 0.325 m

Explanation:

a) The wave length is the distance between 2 crests λ = 6.1m

b) The period of the wave is the time it takes from the lowest point to the next lowest point, which is twice the time it takes from the lowest point to the highest point = 2*2.3 = 4.6 s

c) The speed of the wave is the distance per unit of time, or wave length over period = 6.1 / 4.6 = 1.326 m/s

d)The amplitude A is half the distance from the highest point to the lowest point = 0.65 / 2 = 0.325 m

6 0
3 years ago
The position of a car at time t is given by the function p(t)=t2 2t−4. What is the velocity when p(t)=11? assume t≥0
AysviL [449]

The velocity when function p(t)=11 is 8 .

According to the question

The position of a car at time t  represented by function :

p(t)=t^{2} +2t-4

Now,

When  function p(t) = 11 , t will be

p(t)=t^{2} +2t-4

11 = t²+2t-4

0 = t² + 2t - 15

or

t² +2t-15 = 0

t² +(5-3)t-15 = 0

t² +5t-3t-15 = 0

t(t+5)-3(t+5) = 0

(t-3)(t+5) = 0  

t = 3 , -5  

as t cannot be -ve as given ( t≥0)

so,

t = 3

Now,

the velocity when p(t)=11

As we know velocity = \frac{position}{time}

therefore to get the value of velocity from  function p(t)

we have to differentiate the function with respect to time

\frac{d(p(t))}{dt} =\frac{d}{dt} (t^{2} +2t-4)

v(t) = 2t + 2  

where v(t) = velocity at that time

as t = 3 for  p(t)=11  

so ,

v(t) = 2t + 2  

v(t) = 2*3 + 2  

v(t) = 8

Hence, the velocity when function p(t)=11 is 8 .

To know  more about function here:

brainly.com/question/12431044

#SPJ4

4 0
2 years ago
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