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2 years ago

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Consider an electrophysiology experiment in which a 0.200 ~mm0.200 mm diameter round silver wire is used as a stimulation electr

ode. In this experiment, a current of 4.69~ mA4.69 mA is applied through the wire. The density of free electrons in silver is 5.86 \times 10^{28}~ \frac{\text{electrons}}{m^3}5.86×10 28 m 3 electrons . Calculate the magnitude of the drift velocity of the electrons in the silver wire under these conditions. Enter your answer in units of \frac{\text{meters}}{\text{second}} second meters using calculator scientific notation (e.g., 1.23e-41.23e−4 for 0.0001230.000123).

1 answer:

Kaylis [27]2 years ago

4 0

Explanation:

Below is an attachment containing the solution.

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2 years ago

Read 2 more answers

Am i correct? If not then which one

cupoosta [38]

Answer:

Yes, it's correct

Explanation:

Newton's second Law states that the acceleration of an object is proportional to the net force applied on it, according to the equation:

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where

F is the net force on the object

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a is the acceleration of the object

We can re-arrange the previous equation in order to solve explicitely for a, the acceleration, and we find:

$F=ma\\\frac{F}{m}=\frac{ma}{m}\\\frac{F}{m}=a\\a=\frac{F}{m}$

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3 years ago

A van de Graaff generator accelerates electrons so that they have energies equivalent to that attained by falling through a pote

Marat540 [252]

Answer:

Hello your question is incomplete hence I will give you a general answer on how A van de Graaff generator works

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Explanation:

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2 years ago

Water (density = 1x10^3 kg/m^3) flows at 15.5 m/s through a pipe with radius 0.040 m. The pipe goes up to the second floor of th

RUDIKE [14]

Answer:

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

Explanation:

By assuming that fluid is incompressible and there are no heat and work interaction through the line of current corresponding to the pipe, we can calculate the speed of the water floor in the pipe on the second floor by Bernoulli's Principle, whose model is:

$P_{1} + \frac{\rho\cdot v_{1}^{2}}{2}+\rho\cdot g\cdot z_{1} = P_{2} + \frac{\rho\cdot v_{2}^{2}}{2}+\rho\cdot g\cdot z_{2}$ (1)

Where:

$P_{1}$ , $P_{2}$ - Pressures of the water on the first and second floors, measured in pascals.

$\rho$ - Density of water, measured in kilograms per cubic meter.

$v_{1}$ , $v_{2}$ - Speed of the water on the first and second floors, measured in meters per second.

$z_{1}$ , $z_{2}$ - Heights of the water on the first and second floors, measured in meters.

Now we clear the final speed of the water flow:

$\frac{\rho\cdot v_{2}^{2}}{2} = P_{1}-P_{2}+\rho \cdot \left[\frac{v_{1}^{2}}{2}+g\cdot (z_{1}-z_{2}) \right]$

$\rho\cdot v_{2}^{2} = 2\cdot (P_{1}-P_{2})+\rho\cdot [v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})]$

$v_{2}^{2}= \frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})$

$v_{2} = \sqrt{\frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2}) }$ (2)

If we know that $P_{1}-P_{2} = 0\,Pa$ , $\rho=1000\,\frac{kg}{m^{3}}$ , $v_{1} = 15.5\,\frac{m}{s}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $z_{1}-z_{2} = -3.5\,m$ , then the speed of the water flow in the pipe on the second floor is:

$v_{2}=\sqrt{\left(15.5\,\frac{m}{s} \right)^{2}+2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (-3.5\,m)}$

$v_{2} \approx 13.100\,\frac{m}{s}$

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

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2 years ago