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vredina [299]
3 years ago
10

Which of the following are physical properties of nonmetals? a. they are brittle in the solid form b. most are gases at room tem

perature c. they have low boiling and melting points d. all of the above are physical properties of nonmetals
Physics
2 answers:
iVinArrow [24]3 years ago
7 0

<u>Answer:</u>

<em>All the above properties are the physical properties of non metals. </em>

<u>Explanation:</u>

The elements in the periodic table can be classified as metals, nonmetals and metalloids based on their physical and chemical properties. The nonmetals are basically the elements present in the chalcogenides groups such as oxygen, nitrogen, chlorine, fluorine etc.

For the most part, the nonmetals exist in solid and gaseous stage just with the exception of bromine which is a fluid nonmetal at room temperature. A portion of the nonmetals can likewise exist in the strong state yet those solids will be weak in nature.

<em>So every one of the alternatives is coordinating with the physical properties of nonmetals.</em>

Black_prince [1.1K]3 years ago
6 0

Option (D) is correct that all of the above are physical properties of non metals.

#$# THANK YOU #$#

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A small sphere with mass m is attached to a massless rod of length L that is pivoted at the top, forming a simple pendulum. The
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a) see attached, a = g sin θ

b)

c)   v = √(2gL (1-cos θ))

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In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by

          Wₓ = m a

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b) The diagram is the same, the only thing that changes is the angle that is less

                θ' = 9/2  θ

             

c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.

The easiest way to find linear speed is to use conservation of energy

Highest point

            Em₀ = mg h = mg L (1-cos tea)

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          Emf = K = ½ m v²

          Em₀ = Emf

          g L (1-cos θ) = v² / 2

              v = √(2gL (1-cos θ))

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