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satela [25.4K]
3 years ago
5

Which statements describe characteristics of most metals? Check all that apply.

Physics
2 answers:
RSB [31]3 years ago
7 0

Answer:

Option A, B, D and E are the characteristics of Metal

Explanation:

Some of the common characteristics of most of the metals are -

a) Most of the metal have lustrous surface which means they glitter in the presence of light for example - Iron, copper etc.

b) All metals are malleable which means they can be molded into different shape on beating for example copper can be converted into copper wire, jug, plates etc.

c) All metals are good carrier of charge and thus they are good conductors. These metals have valence shells electron which are free to move with a small force. Good metal conductors are copper , iron etc

d) Most of the metal are solid at room temperature.

DerKrebs [107]3 years ago
6 0

The answer is:

A. They can be formed into wires.

B.They are shiny.

D. They are good conductors

E.can be easily shaped by hammering or pounding.

The explanation:

Let's see the characteristics of the most metals:

1) the most metals can be hit by a hammer and form a thin sheets without breaking and this called malleability.

for example: Aluminium and copper

2) They can form into a very thin wires and this called ductility

for example: silvar , Aluminium and copper.

3) The metal can conduct the heat and the electricity very easy and quick, this mean that the meals are good conductor for the heat and electricity.

4)The metals like gold can be used at jewellery because it is very shiny.

5) and answer C is wrong because most metals are solid at room temperature.

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Answer:

(a)\ F_{No} = [P_{No} - \frac{P_{area}}{2}]* A

(b)\ F_{No} = 771.125N

Explanation:

Given

d_D = 6000ft ---- Altitude of container in Denver

A = 0.0155m^2 -- Surface Area of the container lid

P_D = 79000Pa --- Air pressure in Denver

P_{No} = 100250Pa --- Air pressure in New Orleans

<em>See comment for complete question</em>

Solving (a): The expression for F_{No

Force is calculated as:

F = \triangle P * A

The force in New Orleans is:

F_{No} = \triangle P * A

Since the inside pressure is half the pressure at sea level, then:

\triangle P = P_{No} - \frac{P_{area}}{2}

Where

P_{area} = 101000Pa --- Standard Pressure

Recall that:

F_{No} = \triangle P * A

This gives:

F_{No} = [P_{No} - \frac{P_{area}}{2}]* A

Solving (b): The value of F_{No

In (a), we have:

F_{No} = [P_{No} - \frac{P_{area}}{2}]* A

Where

A = 0.0155m^2

P_{No} = 100250Pa

P_{area} = 101000Pa

So, we have:

F_{No} = [100250 - \frac{101000}{2}] * 0.0155

F_{No} = [100250 - 50500] * 0.0155

F_{No} = 49750* 0.0155

F_{No} = 771.125N

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