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DaniilM [7]
3 years ago
5

A capacitor consists of two concentric cylinders. The inner cylinder has a radius of 0.001 m and the outer cylinder a radius of

0.0011 m. The length of the capacitor is 1 m. If centered on the z-axis, the region 0 < ∅ < π has a dielectric constant of 2 and the region π < ∅ < 2π a dielectric constant of 4. Find the capacitance. Ignore fringing fields.
Physics
1 answer:
AlexFokin [52]3 years ago
5 0

Answer:

The capacitance is 1.75 nF

Explanation:

From the question we are given that

    The inner radius is r_{in}  = 0.001

     The outer radius is r_{out} = 0.0011 \ m

    Length of the capacitor is L = 1m

    The dielectric constant is Di = 2 \ for  \ 0 < \phi < \pi

   The dielectric constant is  Di_2  = 4 \ for \ \pi < \phi < 2\pi

Generally the capacitance of a capacitor can be mathematically represented as

                C = \frac{\pi \epsilon_0 Di_1 L}{ln\frac{r_{out}}{r_{in}} } + \frac{\pi \epsilon_0 Di_2L}{ln\frac{r_{out}}{r_{in}} }

                   = \frac{\pi \epsilon_0 L (Di_1 + Di_2)}{ln\frac{r_{out}}{r_{in}} }

                  = \frac{(3.142)(8.85*10^{-12})(1)(2+4)}{ln\frac{0.0011}{0.001} }

                  =1.75*10^{-9} F

                  1.75nF

                 

                   

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A 4.25 kg block is sent up a ramp inclined at an angle theta=37.5° from the horizontal. It is given an initial velocity ????0=15
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Answer:

d = 11.79 m

Explanation:

Known data

m=4.25 kg  : mass of the block

θ =37.5°  :angle θ of the ramp with respect to the horizontal direction

μk= 0.460  : coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the block on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the block

W: Weight of the block : In vertical direction

N : Normal force : perpendicular to the ramp

f : Friction force: parallel to the ramp

Calculated of the W

W= m*g

W=  4.25 kg* 9.8 m/s² = 41.65 N

x-y weight components

Wx= Wsin θ= 41.65*sin 37.5° = 25.35 N

Wy= Wcos θ =41.65*cos 37.5° =33.04 N

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We apply the formula (1)

∑Fy = m*ay    ay = 0

N - Wy = 0

N = Wy

N = 33.04 N

Calculated of the f

f = μk* N= 0.460*33.04

f = 15.2 N

We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax  ,  ax= a  : acceleration of the block

-Wx-f = m*a

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-40.55 =  (4.25)*a

a = (-40.55)/ (4.25)

a = -9.54 m/s²

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Because the block moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :

vf²=v₀²+2*a*d Formula (2)

Where:  

d:displacement  (m)

v₀: initial speed  (m/s)

vf: final speed   (m/s)

Data:

v₀ = 15 m/s

vf = 0

a = -9.54 m/s²

We replace data in the formula (2)  to calculate the distance along the ramp the block reaches before stopping (d)

vf²=v₀²+2*a*d

0 = (15)²+2*(-9.54)*d

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The energy lost is given by subtracting the final kinetic energy from the initial kinetic energy hence

Energy lost=350-312.5=37.5 KJ

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