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White raven [17]

2 years ago

12

What is elasticity in polymer generally related to?

2 answers:

Andrej [43]2 years ago

6 0

The elasticity of a polymer is primarily due to the structure of the molecule and the cross-linking between strands. Hydrogen bonding is a contributor to the shape of the molecule, but not a major player in terms of elasticity. We would have to answer "false".
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slega [8]2 years ago

4 0

It should be elastomer

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A radioactive isotope of the element potassium decays to produce argon. If the ratio of argon to potassium is found to be 31:1,

iris [78.8K]

Answer:

Explanation:

Argon to potassium ratio after 1 half life = 1:1

After 2 half lives = 75/25= 3:1

After 3 half lives = 87.5/12.5= 7:1

After 4 half lives = 93.75/6.25 = 15:1

After 5 half lives = 96.875/3.125 = 31/1

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You are given two rectangular blocks of shiny metal, Block A and Block B, and are asked to determine which one will float in a b

vladimir2022 [97]

Answer:

Explanation:

Volume of block A = 10 x 6 x 1 = 60 cm³

Mass of block A = 630 g

density of mass A = mass / density

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7 0

3 years ago

A bug flying horizontally at 0.65 m/s collides and sticks to the end of a uniform stick hanging vertically. After the impact, th

irina [24]

The angular momentum is defined as,

$L=I\omega$

Acording to this text we know for conservation of angular momentum that

$L_i=L_f$

Where $L_i$ is initial momentum

$L_f$ is the final momentum

How there is a difference between the stick mass and the bug mass, we define that

Mass of the bug= m

Mass of the stick=10m

At the point 0 we have that,

$L_i=mvl$

Where l is the lenght of the stick which is also the perpendicular distance of the bug's velocity

vector from the point of reference (O), and ve is the velocity

At the end with the collition we have

$L_f=(I_b+I_s)\omega$

Substituting

$L_f=(ml^2+\frac{10ml^2}{3})\omega$

$L_f=\frac{13}{10}ml^2w$

$m(0.65)l=\frac{13}{10}ml^2 \omega$

$\omega=\frac{1}{2l}$

Applying conservative energy equation we have

$\frac{1}{2}(I_b+I_s)\omega^2=mgh+10mgh'$

$\frac{1}{2}(ml^2+\frac{10ml^2}{3})(\frac{1}{2l})^2=mg(l-lcos\theta)+\frac{10}{2}mg(l-lcos\theta)$

Replacing the values and solving

$l=\frac{13}{0.54g}$

Substituting

l=\frac{13}{0.54(9.8)}

$l=2.45cm$

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3 years ago