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

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

Read 2 more answers

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
field trying to move them.

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