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3 years ago

Steel is made of atoms of iron and carbon. Would iron and carbon form metallic bonds? Explain your answer choice.

2 answers:

7 0

Sample Response: "No, steel and carbon would not form metallic bonds because metallic bonds only form between metals. Iron is a metal, but carbon is not."

AlladinOne [14]3 years ago

7 0

Explanation:

Metallic bonds are the bonds that are formed between positively charged atoms in which mobile electrons are shared between the cations.

Also, metallic bonds are formed between metals.

Since, iron is a metal and carbon is a non-metals, therefore, they cannot form metallic bonds.

Thus, we can conclude that no, iron and carbon would not form metallic bonds.

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How large must the coefficient of static friction be between the tires and road, if a car rounds a level curve of radius 85 m at

tatuchka [14]

Answer:

0.66

Explanation:

By using the formula

u = v^2 / r g

Where u is coefficent of friction

u = 23.5 × 23.5 / (85 × 9.8)

u = 0.66

4 0

3 years ago

Please help this is for my final exam

ki77a [65]

Answer:

Explanation:

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3 years ago

A. What is the coefficient of kinetic friction?

Andrews [41]

Explanation:

Coefficient of kinetic friction is the resistive force that opposes the motion of a body as it moves and is in contact with another body.

It is found by dividing the frictional force by the normal force.

Friction is a force that opposes motion.
Static friction is for bodies that are not in motion
Kinetic friction is for moving bodies.

7 0

3 years ago

According to Ohm’s law, determine the experimental current for these values in Table A.

mojhsa [17]

Answer:

0.25

1.0

2.5

Explanation:

7 0

4 years ago

Two objects are dropped from rest from the same height. Object A falls through a distance Da and during a time t, and object B f

stiv31 [10]

Answer:

Da=(1/4)Db

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

When s = Da, t = t

$s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times t^2\\\Rightarrow Da=\frac{1}{2}at^2$

When s = Db, t = 2t

$s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times (2t)^2\\\Rightarrow Db=\frac{1}{2}a4t^2$

Dividing the two equations

$\frac{Da}{Db}=\frac{\frac{1}{2}at^2}{\frac{1}{2}a4t^2}=\frac{1}{4}\\\Rightarrow \frac{Da}{Db}=\frac{1}{4}\\\Rightarrow Da=\frac{1}{4}Db$

Hence, Da=(1/4)Db

3 0

4 years ago