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Sunny_sXe [5.5K]
2 years ago
5

1. Compared to most other metals, what properties do the alkali metals have? А low melting points and high densities B low melti

ng points and low densities C high melting point and high densities D high melting points and low densities​
Physics
1 answer:
ivann1987 [24]2 years ago
3 0

Answer:

B. low melting points and low densities

Explanation:

Alkali metals are any of the monovalent elements found in Group IA of the periodic table. They readily lose their one valence electron to form ionic compounds with nonmetals. Examples of alkali metal are Potassium (K), Lithium (L), and Sodium (Na).

Compared to most other metals, the chemical properties that alkali metals have are low melting points (28.5°C) and low densities that is typically less than 1 grams per cubic centimeters.

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An atom undergoes nuclear decay, but its atomic number is not changed.
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Answer:

A. Gamma decay

Explanation:

A form of nuclear decay in which the atomic number is unchanged is a gamma decay.

The atom has undergone a gamma decay.

In a gamma decay, no changes occur to the mass and atomic number of the substance.

  • Gamma rays have zero atomic and mass numbers.
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A theory was not originally a hypothesis. true or false?
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Which one doesn't belong in the group? oxygen, sulfur, selenium,
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Using Gauss's law, calculate the electric field at a point distance s from a long wire bearing uniform charge density. i need he
11111nata11111 [884]

Answer:

E = 2k  \frac{\lambda}{ r}

Explanation:

Gauss's law states that the electric flux equals the wax charge between the dielectric permeability.

We must define a Gaussian surface that takes advantage of the symmetry of the problem, let's use a cylinder with the faces perpendicular to the line of charge. Therefore the angle between the cylinder side area has the same direction of the electric field which is radial.

            Ф = ∫ E . dA = E ∫ dA = q_{int} /ε₀

tells us that the linear charge density is

            λ = q_ {int} /l

            q_ {int} = l λ

we substitute

            E A = l λ /ε₀

is area of ​​cylinder is

           A = 2π r l

we substitute

            E = \frac{ l \ \lambda}{ \epsilon_o \ 2\pi  \ r \ l }

             E = \frac{\lambda}{ 2\pi  \epsilon_o \ r}

the amount

            k = 1 / 4πε₀

            E = 2k  \frac{\lambda}{ r}

5 0
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