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Contact [7]
3 years ago
9

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 10.5 cm, giving it a cha

rge of -22.3 µC.. . (a) Find the magnitude of the electric field just inside the paint layer. (b) Find the magnitude of the electric field just outside the paint layer. (c) Find the magnitude of the electric field 5.00 cm outside the surface of the paint layer.

Physics
2 answers:
soldier1979 [14.2K]3 years ago
8 0
The formula that will be used in this problem is E = q/ 4pi*r^2 z where z is the elctric charge constant equal to 8.854  *10 ^-12. The magnitude using r equal to 0.0525 m and q equal to -22.3 *10^-6 C is equal to -22.3 *10^-6/ 4pi*(0.0525)^2 *8.854  *10 ^-12 or equal to -7.272 *10 ^7. The magnitude 5 cm outside the surface is -22.3 *10^-6<span>/ 4pi*(0.0525+0.05)^2 *8.854  *10 ^-12 equal to -1.908 *10^7. 

</span>
olya-2409 [2.1K]3 years ago
8 0

Since it is hard to write the mathematics equations here so i would like to share it in a picture

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3 years ago
A block of wood of mass 24 kg floats on water. The volume of the block below the surface of the water and the density of the woo
Slav-nsk [51]

Answer:

0.024m^3

Explanation:

=======

Answer:

=======

Given:-

Mass of the block of wood = 24 kg

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Density of water = 1000kg/m^3

Now,

Density of wood is given by,

\frac{m}{v} = \frac{24}{0.032} \\

\frac{m}{v} = 750 \: kg/m ^{3}

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Since the volume of the wood is equal to the volume of water displaced, it is 0.024m^3

=====

Note:

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=> The volume of the wood below the water surface is the volume of water displaced.

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3 years ago
Question 2 of 10<br> Which two types of energy does a book have as it falls to the floor?
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Two adjacent natural frequencies of an organ pipe are AMT determined to be 550 Hz and 650 Hz. Calculate (a) the M fundamental fr
allochka39001 [22]

Answer:

The fundamental frequency and length of the pipe are 100 Hz and 1.7 m.

Explanation:

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Using formula of organ pipe

f=\dfrac{nv}{2L}

For 550 Hz,

550=\dfrac{n\times340}{2L}...(I)

For 650 Hz,

650=\dfrac{(n+1)\times340}{2L}...(II)

From equation (I) and (II)

550-650=\dfrac{340}{2L}-\dfrac{340}{L}

L=\dfrac{340}{2\times100}

L=1.7\ m

(a). We need to calculate the fundamental frequency for n = 1

Using formula of  fundamental frequency

=f=\dfrac{n\lambda}{2L}

put the value of L

f=\dfrac{1\times340}{2\times1.7}

f=100\ Hz

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3 years ago
Un acróbata de 60 kg realiza un acto de equilibrio sobre un bastón. El extremo del bastón, en contacto con el piso, tiene un áre
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So, the pressure the cane exerts on the floor is 6.39\times 10^6\ Pa.

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3 years ago
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