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Rudiy27
4 years ago
10

An object becomes charged when the atoms in the object gain or lose

Physics
1 answer:
ICE Princess25 [194]4 years ago
5 0
The answer is c......
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Suppose you test a gas in the laboratory. You learn that the gas is made up of carbon atoms and oxygen atoms. Is the gas a compo
Shtirlitz [24]
Maybe, but she hasn't proved it yet. As another example, somebody may bring a jar of gas to your laboratory, and you test the gas in the jar and you find nitrogen atoms, oxygen atoms, argon atoms, and krypton atoms, and there isn't a single compound in it. The gas in the jar is a MIXTURE of gases . . . a mixture that we call "Air".
7 0
3 years ago
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Why must the electric field be normal to the surface at every point of a charged conductor?
solong [7]
-- If the field were inclined to the surface, then it would have
some component parallel to the surface. 

-- Then, since we're talking about a conductor, the charges
on the object would move in response to that component
of the field, until there was no longer any component of the
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3 0
3 years ago
A package of aluminum foil contains 50. ft2 of foil, which weighs approximately 7.5 oz . Aluminum has a density of 2.70 g/cm3. W
Tems11 [23]

Answer:

The thickness of the foil is 0.017 mm.

Explanation:

Given that,

Weight = 7.5 oz = 212.6175 gm

Density = 2.70 g/cm³

Area of aluminium = 50 ft² = 46451.52 cm²

We need to calculate the thickness of the foil

Using formula of density

\rho=\dfrac{m}{V}

\rho=\dfrac{m}{A\times t}

t=\dfrac{m}{A\times \rho}

Where, A = area

t = thickness

m = mass

Put the value into the formula

t=\dfrac{212.6175}{46451.52\times2.70}

t=0.00170\ cm

t=0.017\ mm

Hence, The thickness of the foil is 0.017 mm.

3 0
4 years ago
Learning Goal: To review the concept of conservative forces and to understand that electrostatic forces are, in fact, conservati
CaHeK987 [17]

Explanation:

The electrostatic forces are conservative forces!

The mainly property of the conservative fields is \vec{\nabla} \times \vec E=\vec 0

In spherical coordinates the field's expression is:

\vec E=\frac{Q}{4\pi \epsilon _0 r^2} .\^r

and the curl expression is:

\nabla\times \vec E=\frac{1}{r^2{\sin}\,\theta}\left|\begin{matrix}\hat{r} & r\,\hat{\theta} & r\,{\sin}\,\theta\,\hat{\varphi}  \\& & \\\frac{\partial}{\partial r} & \frac{\partial}{\partial \theta} & \frac{\partial}{\partial \varphi}\\ & & \\E_r & rE_\theta & r{\sin}\,\theta\, E_\varphi\end{matrix}\right|=(0, 0, 0)

to find the expression for the potential function associated:

\vec E=\vec \nabla . V, \Delta V= V_b-V_a=-\int _c \vec E.d\vec l=-\int _c E\^r.dr\^r=-\int _c Edr=\int \limits^a_b \frac{Q}{4\pi \epsilon _0 r^2} dr= \frac{Q}{4\pi \epsilon _0}.(\frac{1}{r}|^b_a)= \frac{Q}{4\pi \epsilon _0}.(\frac{1}{b}-\frac{1}{a})

5 0
3 years ago
Pls answerrrrr thisssss
Jobisdone [24]

The amplitude of wave-c is 1 meter.

The speed of all of the waves is (12meters/2sec)= 6 m/s.

The period of wave-a is 1/2 second.

4 0
4 years ago
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