Name the parts of an atom and how each is charged, either positive, negative or neutral or uncharged

On a smooth horizontal floor an object

lukranit [14]

Answer:

what do you mean by this?

Explanation:

7 0

3 years ago

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While many elemental spectral lines are visible, almost all molecular lines lie in the _____ portion of the spectrum, since they

-Dominant- [34]

Answer:

Infra and Red

Explanation:

While many elemental spectral lines are visible, almost all molecular lines lie in the infra portion of the spectrum, since they are at much lower energy. while many elemental spectral lines are visible, almost all molecular lines lie in the red portion of the spectrum, since they are at much lower energy.

3 0

3 years ago

1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x

allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$

With $(\epsilon_{0})$ been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$

Explanation:

(A) Considering a uniform linear density $λ_{0}$ on the ring, then:

$dQ=\lambda dl$ (1)⇒ $Q=\lambda_{0} 2\pi R$ (2)⇒ $\lambda_{0}=\frac{Q}{2\pi R}$ (3)

Applying the technique of charge integration for finite charges:

$V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r' }} \, dQ$ (4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

$r'=\sqrt{R^{2} +Z^{2}}$ (5)

Using the expressions (1),(4) and (5) you obtain:

$V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}} }} \, d\phi$

Integrating results:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$ (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

$|E| =-\nabla V$ (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

$|E| =-\frac{dV(z)}{dz}$ (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$ (S_b)

4 0

3 years ago

How many million Christmas trees were sold in the year 2000?

Sergio039 [100]

Explanation:

Christmas tree production occurs worldwide on Christmas tree farms, in artificial tree factories and from native strands of pine and fir trees. Christmas trees, pine and fir trees purposely grown for use as a Christmas tree, are grown on plantations in many western nations, including Australia, the United Kingdom and the United States. In Australia, the industry is relatively new, and nations such as the United States, Germany and Canada are among world leaders in annual production.

Great Britain consumes about 8 million trees annually, while in the United States between 35 and 40 million trees are sold during the Christmas season. Artificial Christmas trees are mostly produced in the Pearl River delta area of China. Christmas tree prices were described using a Hotelling-Faustmann model in 2001, the study showed that Christmas tree prices declined with age and demonstrated why more farmers do not price their trees by the foot. In 1993, economists made the first known demand elasticity estimates for the natural Christmas tree market.

5 0

3 years ago

A simple dipole consists of two charges with the same magnitude, q, but opposite sign separated by a distance d. The EDM (electr

DedPeter [7]

Answer:

a. dW = ∫pEsinθdθ b. W = p.E

Explanation:

a. We know torque τ = p × E = pEsinθ where θ is the angle between p and E

Let the torque τ rotate the dipole by an amount dθ. So, the workdone dW = ∫τdθ = ∫pEsinθdθ

b. So, the total work done is gotten by integrating from 90 to θ. So,

W = ∫₉₀⁰dW

= ∫₉₀⁰pEsinθdθ

= pE∫₉₀⁰sinθdθ

= pE(cosθ - cos90)

=pEcosθ

= p.E

8 0

3 years ago