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lilavasa [31]
2 years ago
15

Electric fish generate current with biological cells called electroplaques, which are physiological emf devices. The electroplaq

ues in the South American eel are arranged in 140 rows, each row stretching horizontally along the body and each containing 5000 electroplaques. Each electroplaque has an emf of 0.15 V and an internal resistance of 0.25 Ω .If the water surrounding the fish has a resistance of 800Ω .How much current can the eel produce in water from near its head to near its tail?
Physics
1 answer:
Harlamova29_29 [7]2 years ago
5 0

Answer:

current in water = 0.924 A

Explanation:

Let the current in each row be i.

Thus, current in water is contributed by each row and total current in water becomes 140i.

We are given;

emf of each electroplaque = 0.15 V

Number of electroplaques = 5000

internal resistance = 0.25 Ω

resistance = 800Ω

Applying Kirchoff's Voltage Law to row and water, we have;

5000E − (5000r)i − 800(140i) = 0

Rearranging;

5000E = (5000r)i + 800(140i)

Plugging in the relevant values;

5000 x 0.15 = i((5000 x 0.25) + 112,000)

750 = 113,250i

i = 750/113,250

i = 0.0066 A

Recall earlier, the current in water is 140i.

Thus, current in water = 140 x 0.0066

= 0.924

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If two connected points objects pass through the same set of three points, the shapes created by each will be identical, regardl
NeX [460]

Answer:

This is false

Explanation:

This is why the answer to this question is false. If these types of objects (2 points connected) should pass through same sets of 3 points, irrespective of the order that each object was plotted, we will not get identical shapes created.

The shape that is going to be created will be dependent on the pattern/order that was involved in the plotting. If it was identical, then we will have identical shapes. But if the order during plotting is different then we will have different shapes created.

Thank you!

5 0
3 years ago
An oil tanker has collided with a smaller vessel, resulting in an oil spill in a large, calm-water bay of the ocean. You are inv
PolarNik [594]

Complete Question

An oil tanker has collided with a smaller vessel, resulting in an oil spill in a large, calm-water bay of the ocean. You are investigating the environmental effects of the accident and need to know the area of the spill. The tanker captain informs you that 18000 liters of oil have escaped and that the oil has an index of refraction of n = 1.1. The index of refraction of the ocean water is 1.33. From the deck of your ship you note that in the sunlight the oil slick appears to be blue. A spectroscope confirms that the dominant wavelength from the surface of the spill is 485 nm. Assuming a uniform thickness, what is the largest total area oil slick

Answer:

The  largest total area of the oil slick  A = 8.257 *10^{9} \ m^2

Explanation:

From the question we are told that

     The volume of oil the escaped is  V  = 18000 \ L

    The refractive index of oil is n_o = 1.1

     The refractive index of water is n_w = 1.33

      The wavelength of the light  is \lambda = 485 \ nm =  485 * 10^{-9} \ m

         

Generally the thickness of the oil for condition of constructive interference between the oil and the water is mathematically represented as

          d = m *\frac{\lambda}{2n_w}

Where is the order of interference of the light and it value ranges from 1, 2, 3,...n

It is usually take as 1 unless stated otherwise by the question

substituting value

      d = 1 * \frac{485 *10^{-9}}{2 * 1.1}    

      d = 218 nm    

The are can be mathematically evaluated as

        A = \frac{V}{d}

Substituting values

        A = \frac{18000}{218*10^{-8}}

        A = 8.257 *10^{9} \ m^2

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2 years ago
(i). A ball of mass 1.500 kg is attached to the end of a cord 1.50 m long. The ball moves in a horizontal circle. If the cord ca
Aleks04 [339]

(a) Let v be the maximum linear speed with which the ball can move in a circle without breaking the cord. Its centripetal/radial acceleration has magnitude

a_{\rm rad} = \dfrac{v^2}R

where R is the radius of the circle.

The tension in the cord is what makes the ball move in its plane. By Newton's second law, the maximum net force on it is

F = (1.500\,\mathrm{kg}) a_{\rm rad}

so that

(1.500\,\mathrm{kg}) \dfrac{v^2}{1.50\,\rm m} = 64.0\,\mathrm N

Solve for v :

v^2 = \dfrac{(64.0\,\mathrm N)(1.50\,\mathrm m)}{1.500\,\rm kg} \\\\ \implies \boxed{v = 8.00 \dfrac{\rm m}{\rm s}}

(b) The net force equation in part (a) leads us to the relation

F = \dfrac{mv^2}R \implies v = \sqrt{\dfrac{FR}m}

so that v is directly proportional to the square root of R. As the radius R increases, the maximum linear speed v will also increase, so the cord is less likely to break if we keep up the same speed.

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Alekssandra [29.7K]

Answer: "B" Changing Position

Great Question!

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<u><em /></u>

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