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PolarNik [594]
3 years ago
14

Based in bonding theory, explain why heat capacity increases when you consider metals, ceramics and polymers.

Engineering
1 answer:
Sonbull [250]3 years ago
4 0

Answer:

Metals are good conductors are heat. Many applications of ceramics, such as their use as insulating materials cause higher heating capacity. In the case of polymers, we have to distinguish between the heat capacity of liquid, rubbery and glassy polymers. The heat capacity increases with increasing temperature, therefore, a liquid or rubbery polymer can hold more energy than a solid polymer.  

Explanation:

Valence bond (VB) theory is a chemical bonding theory that explains the chemical bonding between two atoms. Like molecular orbital (MO) theory, it explains bonding using principles of quantum mechanics. According to valence bond theory, bonding is caused by the overlap of half-filled atomic orbitals. The two atoms share each other's unpaired electron to form a filled orbital to form a hybrid orbital and bond together. Sigma and pi bonds are part of valence bond theory.

Bonding, Structure and Properties

  1. The bonding and structure of substances determine their properties.
  2. Bonding is the way the atoms are held together.
  3. The structure is the way the atoms are arranged relative to each other.
  4. There are two major types of structure: giant and molecular.
  5. Giant structures go on indefinitely, whereas molecular structures are made up of groups of atoms.

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Write a method called letterCount that takes two String arguments, one containing some text and the other containing a single le
jenyasd209 [6]

Answer:

I am writing a Python program.  Here is the function letterCount. This function takes two string arguments text and letter and return count of all occurrences of a letter in the text.

def letterCount(text, letter):  

 count = 0  # to count occurrences of letter in the text string

 for char in text:  # loop moves through each character in the text

   if letter == char: # if given letter matches with the value in char

     count += 1  # keeps counting occurrence of a letter in text

 return count # returns how many times a letter occurred in text

   

Explanation:

In order to see if this function works you can check by calling this function and passing a text and a letter as following:

print(letterCount('apples are tasty','a'))

Output:

3

Now lets see how this function works using the above text and letter values.

text = apples are tasty

letter = a

So the function has to compute the occurrences of 'a' in the given text 'apples are tasty'.

The loop has a variable char that moves through each character given in the text (from a of apples to y of tasty) so it is used as an index variable.

char checks each character of the text string for the occurrence of letter a.

The if condition checks if the char is positioned at a character which matches the given letter i.e. a. If it is true e.g if char is at character a of apple so the if condition evaluates to true.

When the if condition evaluates to true this means one occurrence is found and this count variable counts this occurrence. So count increments every time the occurrence of letter a is found in apples are tasty text.

The loop breaks when every character in text is traversed and finally the count variable returns all of the occurrences of that letter (a) in the given text (apples are tasty). As a occurs 3 times in text so 3 is returned in output.

The screen shot of program along with output is attached.

4 0
3 years ago
The particle travels along the path defined by the parabola y=0.5x2, where x and y are in ft. If the component of velocity along
JulsSmile [24]

Answer:

D=41.48 ft

a=54.43\ ft/s^2

Explanation:

Given that

y=0.5 x²                      

Vx= 2 t

We know that

V_x=\dfrac{dx}{dt}

At t= 0 ,x=0  

x=\int V_x.dt

At t= 3 s

x=\int_{0}^{3} 2t.dt

x=[t^2\left\right ]_0^3

x= 9 ft

When x= 9 ft then

y= 0.5 x 9²  ft

y= 40.5 ft

So distance from origin is

x= 9 ft ,y= 40.5 ft

D=\sqrt{9^2+40.5^2} \ ft

D=41.48 ft

a_x=\dfrac{dV_x}{dt}

Vx= 2 t

a_x= 2\ ft/s^2

At t= 3 s , x= 9 ft

y=0.5 x²    

a_y=\dfrac{d^2y}{dt^2}

y=0.5 x²    

\dfrac{dy}{dt}=x\dfrac{dx}{dt}

\dfrac{d^2y}{dt^2}=\left(\dfrac{dx}{dt}\right)^2+x\dfrac{d^2x}{dt^2}

Given that

\dfrac{dx}{dt}=2t

\dfrac{dx}{dt}=2\times 3

\dfrac{dx}{dt}=6\ ft/s

a_y=\dfrac{d^2y}{dt^2}=6^2+9\times 2\ ft/s^2

a_y=54\ ft/s^2

a=\sqrt{a_x^2+a_y^2}\ ft/s^2

a=\sqrt{2^2+54^2}\ ft/s^2

a=54.43\ ft/s^2

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Answer:

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