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goblinko [34]
3 years ago
7

Because blood contains charged ions, moving blood develops a Hall voltage across the diameter of an artery. A large artery with

a diameter of 0.75 cm has a flow speed of 0.9 m/s. If a section of this artery is in a magnetic field of 0.24 T, what is the maximum possible potential difference across the diameter of the artery? Answer in units of mV.
Physics
1 answer:
RoseWind [281]3 years ago
3 0

Answer:

maximum possible potential difference = 1.62 mV

Explanation:

given data

diameter d = 0.75 cm = 0.75 × 10^{-2} m

flow speed v = 0.9 m/s

magnetic field B = 0.24 T

to find out

maximum possible potential difference across the diameter

solution

we know that maximum possible potential difference formula is

Vh = Vd× B × d    .............................1

put here value in equation 1 we get

maximum possible potential difference = 0.9 × 0.24 ×  0.75 × 10^{-2}

solve we get

maximum possible potential difference = 1.62 × 10^{-3}

maximum possible potential difference = 1.62 mV

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A thin, uniformly charged insulating rod has a linear charge density λ = 3 nC/m and lies along the x axis from x = 1m to x = 3m.
Contact [7]

Answer:

A) V_A = 11.93~V

B) The vector definition of E-field is

\vec{E} = -1.13\^x + 2.41\^y

where magnitude is E = 2.66 N/m.

Explanation:

The potential of a uniformly charged rod can be found by the method of integration. We will first choose an infinitesimal part on the rod. We will compute the potential of this part at point A. Then we will integrate this potential over the entire rod.

We will use the following formula for electric potential:

V = \frac{1}{4\pi \epsilon_0}\frac{Q}{r}

Let us choose the infinitesimal part a distance 'x' from the origin. Then the distance between this point and point A is

r = \sqrt{x^2+4^2}

The infinitesimal length is 'dx', and the potential of this length is dV. Let's apply the formula:

dV = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{\sqrt{x^2 + 4^2}}

Here, the charge Q is equal to the charge density multiplied by the length. Q = λdx

Now we have to integrate this infinitesimal potential over the rod:

V = \int\limits^3_1 {dV} \, dx = \frac{1}{4\pi \epsilon_0}\int\limits^3_1 {\frac{\lambda}{\sqrt{x^2 + 16}} \, dx

By using an integral table, this can be calculated:

V = \frac{3\times 10^{-9}}{4\pi\epsilon_0}\ln(|\sqrt{x^2+16}+x|)\left \{ {{x=3} \atop {x=1}} \right. \\V = 11.93~V

B) The electric field can be found by a similar approach, but a different formula:

\vec{E} = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}\^r

Let's apply this formula to the infinitesimal part we have chosen.

dE_x = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{x^2 + 4^2}\cos(\theta)\\dE_y = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{x^2 + 4^2}\sin(\theta)

By the geometry sine and cosine terms can be found:

\sin(\theta) = \frac{4}{\sqrt{x^2+16}}\\\cos(\theta) = \frac{x}{\sqrt{x^2 + 16}}

The x- and y-components of the E-field can be found separately by integrating the infinitesimal parts over the entire rod.

E_x = \int\limits^3_1 {dE_x} \, dx = \frac{\lambda}{4\pi\epsilon_0}\int\limits^3_1 {\frac{x}{(x^2+16)^{3/2}}} \, dx  = 1.13(-\^x)\\E_y = \int\limits^3_1 {dE_y} \, dx = \frac{4\lambda}{4\pi\epsilon_0}\int\limits^3_1 {\frac{1}{(x^2+16)^{3/2}}} \, dx  = 2.41(\^y)

So, the final E-field is

\vec{E} = -1.13\^x + 2.41\^y

The magnitude of the E-field is

E = 2.66 N/m

6 0
3 years ago
What is the mass, in kilograms, of an avogadro's number of people, if the average mass of a person is 150 lb ?
Nikolay [14]
First step is to convert the lb to kg as follows:
1 lb = 0.45 kg
Therefore, 150 lb = 150 x 0.45 = 67.5 kg

Avogadro's number = 6.02 x 10^23

Mass of Avogadro's number of people = 6.02 x 10^23 x 67.5  
                                                              = 4.0635 x 10^25 kg
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A small rock is thrown straight up with initial speed V0 from
bagirrra123 [75]

Answer:

12 feet

Explanation:

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Identify the step of meiosis
Nimfa-mama [501]

Answer:

Since cell division occurs twice during meiosis, one starting cell can produce four gametes (eggs or sperm). In each round of division, cells go through four stages: prophase, metaphase, anaphase, and telophase.

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Define temperature ​
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Answer:

temperature is the degree or intensity of heat present in a substance or object.

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