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1 year ago

9

A book rests on a wooden table is an example of

1 answer:

kykrilka [37]1 year ago

7 0

Answer:

Static friction

Explanation:

When there is contact between two surfaces and there is no movement, there is static friction. So, if we have a book that rests on a wooden table, we have an example of Static friction, where this force avoids the movement of the book.

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Two large, parallel, nonconducting sheets of positive charge face each other. What is at points (a) to the left of the sheets, (

alexira [117]

Answer:

a)The electric Field will be zero at the point between the sheets

b) $E_1=\dfrac{\sigma}{\epsilon_0}$

c) $E_2=\dfrac{\sigma}{\epsilon_0}$

Explanation:

Let $\sigma$ be the surface charge density of the of the non conducting parallel sheet.Let consider a Gaussian surface in the form of of cylinder such that its cross-sectional is A . Then there will be flux only due to cross sectional area as the curved sectional is perpendicular to the the electric field so the Electric Flux due to it is zero.

Now using Gauss law we have, E be the electric Field at the distance r from the sheet then

$E\times 2A=\dfrac{\sigma A}{\epsilon_0}\\E=\dfrac{\sigma}{2\epsilon_0}$

The Field will be away from the sheet and perpendicular to it.

a) The Electric Field between them

$E_1=\dfrac{\sigma}{2\epsilon_0}-\dfrac{\sigma}{2\epsilon_0}\\=0$

b)The Electric Field to the right of the sheets

$E_1=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}\\=\dfrac{\sigma}{\epsilon_0}$

c)The Electric Field to the left of the sheets

$E_2=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}\\=\dfrac{\sigma}{\epsilon_0}$

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lara [203]

I think the answer is B

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nexus9112 [7]

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When a 10 V battery is connected to a resistor, 5 A of current flows through the resistor. What is the resistor's value?

SashulF [63]

Given data

*The value of battery voltage is V = 10 V

*The current flows through the resistor is I = 5 A

The formula for the resistor is given by the Ohm's law as

$R=\frac{V}{I}$

Substitute the values in the above expression as

$\begin{gathered} R=\frac{10}{5} \\ =2\text{ ohm} \end{gathered}$

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1 year ago