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antoniya [11.8K]

3 years ago

12

A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and

the outer is positively charged; the magnitude of the charge on each is 18.5pC . The inner cylinder has a radius of 0.350mm , the outer one has a radius of 7.20mm , and the length of each cylinder is 17.0cmPart AWhat is the capacitance? Use 8.854

1 answer:

Musya8 [376]3 years ago

5 0

Answer: 3.12 * 10^12 F ( 3.12 pF)

Explanation: To calculate this capacitor of two hollow, coaxial, iron cylinders, we have to determine the potental differente between them and afeter that to use C=Q/ΔV

The electric field in th eregion rinner<r<router

By using the Gaussian law

∫E*ds=Q inside/εo

E*2*π*rinner^2*L= Q /εo

E=Q/(2*π*εo*r^2)

[Vab]=\int\limits^a_b {E} \, dr

where a and b are the inner and outer radii.

Then we have:

ΔV= 2*k*(Q/L)* ln (b/a)

replacing the values and using that C=Q/ΔV

we have:

C= L/(2*k*ln(b/a)=0.17/(2*9*10^9*3.023)=3.12 pF

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Explanation:

From the given information,

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The mass of the water accumulated in the bucket after 3.20s is:

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To determine the weight of the water accumulated in the bucket, we have:

$W_w = m_w g$

$W_w = ( 0.64 \ kg )(9.80\ m \ /s^2)$

$W_w = 6.272 \ N$

For the speed of the water before hitting the bucket; we have:

$v = \sqrt{2gh}$

$v = \sqrt{2*9.80 \ m/s^2 * 3.60 \ m}$

v = 8.4 m/s

Now, the force required to stop the water later when it already hit the bucket is:

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$F = (8.4 \ m/s)( 0.200 \ L/s)$

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Answer:

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Answer:

Part(a): The equation of motion is $\bf{x(t) = \dfrac{7}{5}~e^{-2t} - \dfrac{2}{5}~e^{-7t}}$ .

Part(b): The equation of motion is $\bf{x(t) = -e^{-2t} + \dfrac{11}{5}~e^{-7t}}$ .

Explanation:

If ' $m$ ' be the mass of the object, ' $k$ ' be the force constant and ' $\beta$ ' be the damping constant, then the equation of motion of the particle can be written as

$\dfrac{d^{2}x}{dt^{2}} + \dfrac{\beta}{m} \dfrac{dx}{dt} + \dfrac{k}{m}x= 0.........................................(I)$

Given $m$ = 1 Kg, $k$ = 14 $N~m^{-1}$ , $\beta$ = 9. Substituting these values in equation (I),

$\dfrac{d^{2}x}{dt^{2}} + 9~\dfrac{dx}{dt} + 14~x= 0$

Taking a trial solution $x(t) = e^{mt}$ , the auxiliary equation can be written as

$m^{2} + 9m + 14 = 0............................................................(II)$

and its solutions are $m_{1} = -2~and~m_{2} = -7$ , resulting the general solution

$x(t) = C_{1}~e^{-2t} + C_{2}~e^{-7t}....................................................................(III)$

The velocity at any instant of time of the mass is

$v(t) = -2C_{1}~e^{-2t} _7~C_{2}~e^{-7t}..............................................................(IV)$

Part(a):

Given $x(t=0) = 1 m,~and~v(t=0) = 0~m~s^{-1}$ . Substituting these values in equation (III) and (IV),

$&& 1 = C_{1} + C_{2}......................(V)\\&and,& 0 = -2C_{1} - 7C_{2}.......................(VI)$

Solving equations (V) and (VI), we have

$C_{1} = \dfrac{7}{5}~and~C_{2} = \dfrac{-2}{5}$

So the equation of motion is

$x(t) = \dfrac{7}{5}~e^{-2t} - \dfrac{2}{5}~e^{-7t}$

Part(b):

Given $x(t=0) = 1 m,~and~v(t=0) = - 12~m~s^{-1}$ . Substituting these values in equation (III) and (IV),

$&& 1 = C_{1} + C_{2}......................(VII)\\&and,& -12 = -2C_{1} -7C_{2}.......................(VIII)$

Solving equations (V) and (VI), we have

$C_{1} = -1~and~C_{2} = \dfrac{11}{5}$

So the equation of motion is

$x(t) = -e^{-2t} + \dfrac{11}{5}~e^{-7t}$

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