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Lisa [10]
3 years ago
12

Elaborate on the classification of aluminum. A) Aluminum is a noble gas and has a full valence electron shell, is chemically non

-reactive, colorless, odorless, and is a gas at room temperature. B) Aluminum is a metal and is shiny, malleable, ductile, conducts heat and electricity, forms basic oxides, and forms cations in aqueous solution. C) Aluminum is a nonmetal and is dull, brittle, poor conductor of heat and electricity, forms acidic oxides, and forms anions in aqueous solution. D) Aluminum is a metalloid and has properties of both metals and nonmetals - it appears lustrous, but is brittle and a poor conductor of heat and electricity.
Physics
2 answers:
guajiro [1.7K]3 years ago
5 0

Elaborate on the classification of aluminum.

                                         *correct answer*


B) Aluminum is a metal and is shiny, malleable, ductile, conducts heat and electricity, forms basic oxides, and forms cations in aqueous solution.

                                                *yw*

scZoUnD [109]3 years ago
4 0
The correct option is (B) <span>Aluminum is a metal and is shiny, malleable, ductile, conducts heat and electricity, forms basic oxides, and forms cations in aqueous solution.

Since Aluminium is in group 13, and all the elements in group 13 are either metals or metalloids(Boron). Hence we are left with option (B) and (D). Boron is the only metalloid in group 13 and aluminium is a metal(not a metalloid); therefore, we are left with only one option which is Option (B). And Aluminium is </span>shiny, malleable, ductile, conducts heat and electricity, forms basic oxides, and forms cations in aqueous solution.<span>
</span><span>
</span>
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Answer:

The <em>net gravitational force it exerts</em> is F_{net}=9.66*10^{-8}N

Explanation:

Newton's Law of Gravitation can be written as

F=\frac{Gm_{1}m_{2}}{r^{2} }

where <em>G is the Gravitational Constant, m1 and m2 are the masses of two objects, and r is the distance between them</em>. In this case, the spheres are loacted in straight line, so instead of a vector r, we have a distance x in meters. The distances and masses are given in the problem, and the smaller sphere is between the other two spheres. This means <u>the sphere 1 is in the middle, the sphere 2 is on the left of 1, and the sphere 3 is on the right of 1</u>, so

F_{21} =\frac{Gm_{1}m_{2}}{x_{21}^{2} } is the force that 2 feels because of 1, and

F_{31} =\frac{Gm_{1}m_{3}}{x_{31}^{2} } is the force that 3 feels because of 1.

<em>If we replace the data in those previous equations</em>, we have that

F_{21} =\frac{G(2.5)(5) }{(0.08)^{2} }=1.3*10^{-7}N

F_{31} =\frac{G(2.5)(8) }{(0.2)^{2} }=-3.34*10^{-8}N

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Answer:

a)   r = (0.6 i- 2039 j ^ + 0.102 k⁾ m  and b) vₓ = 30.0 m / s , v_{y} = 2.04 10⁵ m / s   c) v_{z}  = 1.02 10⁻¹m / s

Explanation:

a) To find the position of the particle at a given moment we must know the approximation of the body, use Newton's second law to find the acceleration

         Fe + Fm = m a

         a = (Fe + Fm) / m

the electric force is

         Fe = q E   k ^

         Fe = 4.25 10-4 60 k ^

         Fe = 2.55 10-2 k ^

the magnetic force is

         Fm = q v x B

         Fm = 4.25 10⁻⁴  \left[\begin{array}{ccc}i&j&k\\30&0&0\\0&0&49\end{array}\right]

         fm = 4.25 10⁻⁴ (-j ^ 30 4)

         fm 0 = ^ -5,10 10⁻² j

We look for every component of acceleration

X axis

      aₓ = 0

there is no force

Axis y

      ay = -5.10 10²/5 10⁻⁵ j ^

      ay = -1.02 107 j ^ / s2

z axis

      az = 2.55 10⁻² / 5 10⁻⁵ k ^

      az = 5.1 10² k ^ m / s²

Having the acceleration in each axis we can encocoar the position using kinematics

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the initial velocity is vo = 30 m / s and an initial position xo = 0

           x = vo t + ½ aₓ t₂2

           x = 30 0.02 + 0

           x = 0.6m

       

Axis y

acceleration is ay = -1.02 10⁷ m / s², a starting position of i = 1m

           y = I + go t + ½ ay t²

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z axis

acceleration is aza = 5.1 10² m / s², the position and initial speed are zero

          z = zo + v₀ t + ½ az t²

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          z = 1.02 10⁻¹ m k ^

therefore the position of the bodies is

   r = (0.6 i- 2039 j ^ + 0.102 k⁾ m

b) x axis

 since there is no acceleration the speed remains constant

          vₓ = 30.0 m / s

Axis y

  let's use the equation v = v₀ + a_{y} t

         v_{y} = 0 + -1.02 10⁷ 0.02

          v_{y} = 2.04 10⁵ m / s

z axis

          v_{z} = vo + az t

          v_{z} = 0 + 5.1 10² 0.02

          v_{z}  = 1.02 10⁻¹m / s

8 0
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