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2 years ago

8

A charge Q is distributed uniformly along the x axis from x1 to x2. What would be the magnitude of the electric field at x0 on t

he x-axis? Assume that ke = 1 4 π ǫ0 and x0 > x2 > x1 for a Coulomb constant of 8.98755 × 109 N · m2 /C 2

1 answer:

Lena [83]2 years ago

8 0

Answer:

E = k Q 1 / (x₀-x₂) (x₀-x₁)

Explanation:

The electric field is given by

dE = k dq / r²

In this case as we have a continuous load distribution we can use the concept of linear density

λ= Q / x = dq / dx

dq = λ dx

We substitute in the equation

∫ dE = k ∫ λ dx / x²

We integrate

E = k λ (-1 / x)

We evaluate between the lower limits x = x₀- x₂ and higher x = x₀-x₁

E = k λ (-1 / x₀-x₁ + 1 / x₀-x₂)

E = k λ (x₂ -x₁) / (x₀-x₂) (x₀-x₁)

We replace the density

E = k (Q / (x₂-x₁)) [(x₂-x₁) / (x₀-x₂) (x₀-x₁)]

E = k Q 1 / (x₀-x₂) (x₀-x₁)

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Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to t

yarga [219]

Answer

The rate at which the magnetic field is changing is $[\frac{dB}{dt} ] = 0.000467 T/s$

Explanation

From the question we are told that

The electric field strength is $E = 3.5mV/m = 3.5 *10^{-3} \ V/m$

The radius is $r = 1.5 \ m$

The rate of change of the magnetic field is mathematically represented as

$\frac{d \phi }{dt} = \int\limits^{} {E \cdot dl}$

Where $dl$ is change of a unit length

$\frac{d \phi}{dt} = A * \frac{dB}{dt}$

Where A is the area which is mathematically represented as

$A = \pi r^2$

So

$E \int\limits^{} { dl} = ( \pi r^2) (\frac{dB}{dt} )$

$E L = ( \pi r^2) (\frac{dB}{dt} )$

where L is the circumference of the circle which is mathematically represented as

$L = 2 \pi r$

So

$E (2 \pi r ) = (\pi r^2 ) [\frac{dB}{dt} ]$

$E = \frac{r}{2} [\frac{dB}{dt} ]$

$[\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }$

substituting values

$[\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }$

$[\frac{dB}{dt} ] = 0.000467 T/s$

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An engineer is working on a new kind of wire product that is about the size of a bacterium cell. Which kind of technology is the

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Microtechnology i hope this will help

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You are participating in a NASA traineeship, working with a group planning a new landing on Mars. Your supervisor has come up wi

aivan3 [116]

Answer:

$h=17005.8 km$

Explanation:

Newton's law of universal gravitation states that the force experimented by a satellite of mass m orbiting Mars, which has mass $M=6.39\times10^{23} kg$ at a distance r will be:

$F=\frac{GMm}{r^2}$

where $G=6.67\times10^{-11}Nm^2/kg^2$ is the gravitational constant.

This force is the centripetal force the satellite experiments, so we can write:

$F=ma_{cp}=mr\omega^2=mr(\frac{2\pi}{T})^2=\frac{4\pi^2mr}{T^2}$

Putting all together:

$\frac{GMm}{r^2}=\frac{4\pi^2mr}{T^2}$

which means:

$r=\sqrt[3]{\frac{GM}{4\pi^2}T^2}$

Which for our values is:

$r=\sqrt[3]{\frac{(6.67\times10^{-11}Nm^2/kg^2)(6.39\times10^{23} kg)}{4\pi^2}(1.026\times24\times60\times60s)^2}=20395282m=20395.3km$

Since this distance is measured from the center of Mars, to have the height above the Martian surface we need to substract the radius of Mars R=3389.5 km , which leaves us with:

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Flow rate = 220*0.355 l/m = 78.1 l/min = 1.3 l/s = 0.0013 m^3/s

Point 2:
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V2 = Flow rate/A2 = 0.0013/0.0008 = 1.625 m/s
P1 = 152 kPa = 152000 Pa

Point 1:
A1 = 2 cm^2 = 0.0002 m^2
V1 = Flow rate/A1 = 0.0013/0.0002 = 6.5 m/s
P1 = ?
Height = 1.35 m

Applying Bernoulli principle;
P2+1/2*V2^2/density = P1+1/2*V1^2/density +density*gravitational acceleration*height
=>152000+0.5*1.625^2*1000=P1+0.5*6.5^2*1000+1000*9.81*1.35
=> 153320.31 = P1 + 34368.5
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Change in time = d(t)
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t2 = 5.00 s
d(t) = (5.00 - 0.00)s = 5.00s

Average acceleration = d(v) / d(t)
(-28.0 m/s) / (5.00 s)
(-28.0 m)/s * 1 / (5.00 s) = -5.60 m/s² to the right

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2 years ago

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