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Sloan [31]
3 years ago
13

A spring has a spring constant of 113 N/m . How much elastic potential energy is stored in the spring when it is compressed by 0

.03 m?A:0.23 j B:0.022 j C:0.051 j D: 1.70 j
Physics
2 answers:
andrezito [222]3 years ago
7 0

Answer:

= 1.7 Joules

Explanation:

The electrical potential energy is given by;

P.E = 1/2 ke²

Where, k is the spring constant, e is the extension

Therefore;

Electrical potential energy = 1/2 × 113 N/m × 0.03 m

                                           =  1.695 Joules

                                           <u>= 1.7 J</u>

Alexus [3.1K]3 years ago
4 0

Answer:

0.051 J

Explanation:

K = 113 N/m

y = 0.03 m

The elastic potential energy of the system is given by

U = 1/2 K y^2 = 0.5 x 113 x 0.03 x 0.03 = 0.051 J

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A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate
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Answer:

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Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and '\epsilon_{0}' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm

Part(c):

If we apply Gauss' law of electrostatics, then

&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}

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