A cat pushes a porcelain statue off a bookshelf with a speed of 0.5 m/s and it smashed on the floor 0.85 sec later.

What is elasticity in polymer generally related to?

Andrej [43]

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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6 0

3 years ago

Read 2 more answers

The color that an object appears to be depends on the select one:

ddd [48]

B is the right answer

3 0

3 years ago

Why is streak generally a more useful property than color in identifying a mineral?

poizon [28]

You may confuse two minerals by their colors so using streak is another way for showing their "true colors "

3 0

3 years ago

La tensión en newtons necesaria para que una onda transversal cuya longitud de onda es 3.33 cm vibre a razón de 625 ciclos por s

NemiM [27]

Answer:

9.34 N

Explanation:

First of all, we can calculate the speed of the wave in the string. This is given by the wave equation:

$v=f \lambda$

where

f is the frequency of the wave

$\lambda$ is the wavelength

For the waves in this string we have:

$f=625 Hz$ , since it completes 625 cycles per second

$\lambda=3.33 cm = 0.033 m$ is the wavelength

So the speed of the wave is

$v=(625)(0.0333)=20.6 m/s$

The speed of the waves in a string is related to the tension in the string by

$v=\sqrt{\frac{T}{\mu}}$ (1)

where

T is the tension in the string

$\mu=\frac{m}{L}$ is the linear density

In this problem:

$m=16.5 g = 16.5\cdot 10^{-3} kg$ is the mass of the string

L = 0.75 m is the its length

Solving the equation (1) for T, we find the tension:

$T=\mu v^2 = \frac{m}{L} v^2 = \frac{16.5\cdot 10^{-3}}{0.75}(20.6)^2=9.34 N$

8 0

3 years ago

A hollow cylinder that is rolling without slipping is given a velocity of 5.0 m/s and rolls up an incline to a vertical height o

inysia [295]

Answer:

The hollow cylinder rolled up the inclined plane by 1.91 m

Explanation:

From the principle of conservation of mechanical energy, total kinetic energy = total potential energy

$M.E_T = \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2 + mgh$

The total energy at the bottom of the inclined plane = total energy at the top of the inclined plane.

$\frac{1}{2}mv_i^2 + \frac{1}{2} I \omega_i^2 + mg(0) = \frac{1}{2}mv_f^2 + \frac{1}{2} I \omega_f^2 + mgh$

moment of inertia, I, of a hollow cylinder = ¹/₂mr²

substitute for I in the equation above;

$\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}mr^2 \omega_i^2) = \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}mr^2 \omega_f^2) + mgh\\\\ but \ v = r \omega\\\\\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}m v_i^2 ) = \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}m v_f^2) + mgh\\\\\frac{1}{2}mv_i^2 +\frac{1}{4}mv_i^2 = \frac{1}{2}mv_f^2 +\frac{1}{4}mv_f^2 +mgh\\\\\frac{3}{4}mv_i^2 = \frac{3}{4}mv_f^2 +mgh\\\\mgh = \frac{3}{4}mv_i^2 - \frac{3}{4}mv_f^2\\\\gh = \frac{3}{4}v_i^2 - \frac{3}{4}v_f^2\\\\$

$h = \frac{3}{4g}(v_1^2 -v_f^2)$

given;

v₁ = 5.0 m/s

vf = 0

g = 9.8 m/s²

$h = \frac{3}{4g}(v_1^2 -v_f^2) =\frac{3}{4*9.8}(5^2 -0) = 1.91 \ m$

Therefore, the hollow cylinder rolled up the inclined plane by 1.91 m

5 0

3 years ago