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ella [17]
3 years ago
6

An electric eel (Electrophorus electricus) can produce a shock of up to 600 V and a current of 1 A for a duration of 2 ms, which

is used for hunting and self-defense. To perform this feat, approximately 80% of its body is filled with organs made up by electrocytes. These electrocytes act as self-charging capacitors and are lined up so that a current of ions can easily flow through them.
a) How much charge flows through the electrocytes in that amount of time?
b) If each electrocyte can maintain a potential of 100 mV, how many electrocytes must be in series to produce the maximum shock?
c) How much energy is released when the electric eel delivers a shock?
d) With the given information, estimate the equivalent capacitance of all the electrocyte cells in the electric eel.
Physics
1 answer:
Irina-Kira [14]3 years ago
8 0

Answer:

2\times 10^{-3}\ C

6000

1.2 J

3.33\times 10^{-6}\ F

Explanation:

I = Current = 1 A

t = Time = 2 ms

n = Number of electrocyte

V = Voltage = 100 mV

Charge is given by

Q=It\\\Rightarrow Q=1\times 2\times 10^{-3}\\\Rightarrow Q=2\times 10^{-3}\ C

The charge flowing through the electrocytes in that amount of time is 2\times 10^{-3}\ C

The maximum potential is given by

V_m=nV\\\Rightarrow n=\dfrac{V_m}{V}\\\Rightarrow n=\dfrac{600}{100\times 10^{-3}}\\\Rightarrow n=6000

The number of electrolytes is 6000

Energy is given by

E=Pt\\\Rightarrow E=V_mIt\\\Rightarrow E=600\times 1\times 2\times 10^{-3}\\\Rightarrow E=1.2\ J

The energy released when the electric eel delivers a shock is 1.2 J

Equivalent capacitance is given by

C_e=\dfrac{Q}{V_m}\\\Rightarrow C_e=\dfrac{2\times 10^{-3}}{600}\\\Rightarrow C_e=3.33\times 10^{-6}\ F

The equivalent capacitance of all the electrocyte cells in the electric eel is 3.33\times 10^{-6}\ F

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ArbitrLikvidat [17]

Answer:

The magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 1.01 x 10^{-4} N/C

Explanation:

given information,

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E = F/q

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3 years ago
A vector is 0.888 m long and points in a 205 degree direction.
Dmitry_Shevchenko [17]

Answer:

-0.805 m

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In this problem we have

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(Missing figure is here: https://www.physicsforums.com/attachments/ch05-p070-jpg.149243/ )

Let's call m_1 = 1.0 kg and m_2=2.0 kg the masses of the two blocks. We can write Newton's second law for both blocks (sum of all forces acting on the block = ma). On block 1, we have two forces: the weight m_1 g pointing downwards and the tension of the string T poiting upwards. On block 2, we have the tension of the string going right and the friction \mu m_2 g going left. Therefore

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