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kirill115 [55]
3 years ago
13

A 180 cm length of string has a mass of 5.0 g. it is stretched with a tension of 8.6 n between fixed supports. (a) what is the w

ave speed for this string
Physics
1 answer:
Svetlanka [38]3 years ago
5 0
P               U               S                 S                 Y


<span>Joy is planning to purchase a sweater that costs $30 dollars at her local department store. The sweaters are on sale for 20% off. Which steps are needed to find the sale price of the sweater?</span>
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While eating a peanut butter and jelly sandwich peanut butter drips on Dana shirt she tries to remove the peanut butter from her
MakcuM [25]

Answer:

The water does not remove the peanut butter because the peanut butter is thicker substance than the water is, therefore it would not remove immediately.. It could also be because the brown dye or color in the peanut butter color or the peanut butter itself has already gone through her clothing Dana would have to remove it as much as she could with a towel water and some soap then throw it to the washer afterward.

Hope this helps, good luck:)

6 0
3 years ago
A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate
Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is \bf{2.81 \times 10^{4}~n~C^{-1}}.

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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It’s a chemical change
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Answer: Refraction

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