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3 years ago

Beth made the chart shown.

Physics

2 answers:

Andru [333]3 years ago

8 0

Answer: A

X: Use DC

Y: Use AC

Explanation:

A on edge

Flauer [41]3 years ago

6 0

Answer:

X: Use DC

Y: Use AC

Explanation:

The objects: Wall clock, T V remote control, Portable music player, are normally powered by batteries, which run DC current.

On the other hand, the objects: Refrigerator, Table lamp, Hair dryer, demand much larger power, and are normally run from the AC plug on the wall.

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The frequency of the vibration is the frequency of the AC supply.

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As a result, the total energy in a ____ system (in other words, a system with no external forces) will remain constant

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Answer:

Isolated or Closed system, both are correct

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3 years ago

What is the strength of the electric field 0.1 mmmm below the center of the bottom surface of the plate

Aleksandr [31]

Complete Question

A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?

Answer:

The values is $E =248.2 \ N/C$

Explanation:

From the question we are told that

The diameter is $d = 12 \ cm = 0.12 \ m$

The charge is $Q = 4.4 nC = 4.4 *10^{-9} \ C$

The distance from the center is $k = 0.1 \ mm = 1*10^{-4} \ m$

Generally the radius is mathematically represented as

$r = \frac{d}{2}$

=> $r = \frac{0.12}{2}$

=> $r = 0.06 \ m$

Generally electric field is mathematically represented as

$E = \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 + k^2 } } ]$

substituting values

$E = \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 + (1.0*10^{-4})^2 } } ]$

$E =248.2 \ N/C$

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3 years ago

F = 2.0*10^20 N,

natka813 [3]

$F=\dfrac{Gm_1m_2}{r^2}$

With the given values of $F,G,m_1,r$ , we have

$2.0\times10^{20}\,\mathrm N=\dfrac{\left(6.67\times10^{-11}\,\frac{\mathrm{Nm}^2}{\mathrm{kg}^2}\right)\left(5.98\times10^{24}\,\mathrm{kg}\right)m_2}{\left(3.8\times10^8\,\mathrm m\right)^2}$