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zaharov [31]
4 years ago
8

A long cylindrical shell (radius = 2.0 cm) has a charge uniformly distributed on its surface. If the magnitude of the electric f

ield at a point 8.0 cm radially outward from the axis of the shell is 85 N/C, how much charge is distributed on a 2.0-m length of the charged cylindrical surface?
Physics
1 answer:
Ahat [919]4 years ago
4 0

Answer:

Q= 7.566 \times 10 ^{-10} \, C

Explanation:

Applying Gauss' Law to a cylindrical shell of radius 8 cm and height h, concentric to the charged shell, we get:

E(r) \cdot 2 \pi r  h= \cfrac{\lambda h}{\epsilon_o}

Where \lambda is the charge per unit length, and so \lambda h = Q is the charge inside the shell, and if we set h=2\, m we can get the answer to our question.

Solving for Q we get:

Q= \epsilon_o E(r) 2 \pi r h

plugging in the values ( in SI units) we get:

Q= 7.566 \times 10 ^{-10} \, C

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Explanation:

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What is the force the results from a 3,000kg car accelerating at 20.00 m/s^2?
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F = ma  = 3000 * 20 = 60 000 N

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A person on a road trip drives a car at different constant speeds over several legs of the trip. She drives for 10.0 min at 50.0
irakobra [83]
<h2>The average speed for the entire trip is 47.5 m/s .</h2>

It is given that for different time span car have different speed and also the person spend 40\ min=\dfrac{40}{60}\ hrs =0.67\ hrs\ . in eating lunch and buying gas.

We know , average speed is total distance covered by total time taken .

Therefore , average speed , v=\dfrac{total\ distance }{total\ times}

v=\dfrac{\dfrac{10}{60}\times 50+\dfrac{19}{60}\times 100+\dfrac{60}{60}\times 55}{\dfrac{10}{60}+\dfrac{10}{60}+\dfrac{60}{60}+ \dfrac{40}{60}}\\\\\\v=47.5\ m/s

Hence, this is the required solution.

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Swings are usually made out of materials that do not conduct heat easily, so that children will not burn themselves as they play
Irina18 [472]

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8 0
3 years ago
The distance between the first and fifth minima of a single-slit diffraction pattern is 0.350 mm with the screen 39.0 cm away fr
rjkz [21]

Answer: a) the width is 2.54 mm

b) the angle is 0.012°

Explanation:

we have the equation:

λ = zw/(L*m)

Where is the distance for the middle of the screen, L is the distance to the screen, w is the widht of the slit and λ is the wavelength.

now we can isolate z and get:

z = λL*m/w

and the distance between the 5 minimum and the first one is 0.035mm

then:

0.035mm = λL*5/w - λL*1/w = λL*4/w

now we can solve it for w.

0.35mm = 570nm*39.0cm*4/w

now, we have all diferent units, lets use nanometers.

1cm = 1x10^-2 m

1mm = 1x10^-3 m

1nm = 1x10^-9m

then: 1 cm = 1x10^7 nm

         1 mm = 1x10^6 nm

then we have:

0.35x10^6 nm = 570nm*39.0x10^7 nm*4/w

w = (570nm*39.0x10^7 nm*4)/0.35x10^6 nm =(570nm*39.0*10*4)/0.35

w = 2540571.4 nm

it is a bigg number, let's write it in milimeters:

w = (2540571.4/10^6) mm = 2.54 mm

the first minimum can be obtained by the equation w*sinθ = mλ by using m = 1

then we have:

2540571.4 nm*sinθ = 570 nm

sinθ = 570/2540571.4 = 0.000224

θ = asin(0.00022) = 0.012°

5 0
3 years ago
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