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3 years ago

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Suppose you take a piece of hard wax from an unlit candle. After you roll the wax between your fingers for a while, it becomes s

oft. What state of matter is the softer wax? Is it possible for the wax to change without changing its state?

1 answer:

romanna [79]3 years ago

5 0

Answer:

An amorphous solid does not have a definite melting point; instead, it melts gradually over a range of temperatures, because the bonds do not break all at once. This means an amorphous solid will melt into a soft, malleable state (think candle wax or molten glass) before turning completely into a liquid.

Explanation:

Hope this helps

May I get braineist pls?

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3. (a) Sate any three properties of an ideal gas as assumed by the kinetic theory of gas.

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Explanation:

The simplest kinetic model is based on the assumptions that: (1) the gas is composed of a large number of identical molecules moving in random directions, separated by distances that are large compared with their size; (2) the molecules undergo perfectly elastic collisions (no energy loss) with each other and with the ...

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2 years ago

The ____________________ of your voice refers to its intensity or loudness.

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That's the amplitude.

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3 years ago

A plane electromagnetic wave, with wavelength 4.1 m, travels in vacuum in the positive direction of an x axis. The electric fiel

marusya05 [52]

(a) $7.32\cdot 10^7 Hz$

The frequency of an electromagnetic waves is given by:

$f=\frac{c}{\lambda}$

where

$c=3.0\cdot 10^8 m/s$ is the speed of light

$\lambda=4.1 m$ is the wavelength of the wave in the problem

Substituting into the equation, we find

$f=\frac{3.0\cdot 10^8 m/s}{4.1 m}=7.32\cdot 10^7 Hz$

(b) $4.60\cdot 10^8 rad/s$

The angular frequency of a wave is given by

$\omega = 2\pi f$

where

f is the frequency

For this wave,

$f=7.32\cdot 10^7 Hz$

So the angular frequency is

$\omega=2\pi(7.32\cdot 10^7 Hz)=4.60\cdot 10^8 rad/s$

(c) $1.53 m^{-1}$

The angular wave number of a wave is given by

$k=\frac{2\pi}{\lambda}$

where

$\lambda$ is the wavelength of the wave

For this wave, we have

$\lambda=4.1 m$

so the angular wave number is

$k=\frac{2\pi}{4.1 m}=1.53 m^{-1}$

(d) $1.03\cdot 10^{-6}T$

For an electromagnetic wave,

$E=cB$

where

E is the magnitude of the electric field component

c is the speed of light

B is the magnitude of the magnetic field component

For this wave,

E = 310 V/m

So we can re-arrange the equation to find B:

$B=\frac{E}{c}=\frac{310 V/m}{3\cdot 10^8 m/s}=1.03\cdot 10^{-6}T$

(e) z-axis

In an electromagnetic wave, the electric field and the magnetic field oscillate perpendicular to each other, and they both oscillate perpendicular to the direction of propagation of the wave. Therefore, we have:

- direction of propagation of the wave --> positive x axis

- direction of oscillation of electric field --> y axis

- direction of oscillation of magnetic field --> perpendicular to both, so it must be z-axis

(f) 127.5 W/m^2

The time-averaged rate of energy flow of an electromagnetic wave is given by:

$I=\frac{E^2}{2\mu_0 c}$

where we have

E = 310 V/m is the amplitude of the electric field

$\mu_0$ is the vacuum permeability

c is the speed of light

Substituting into the formula,

$I=\frac{(310 V/m)^2}{2(4\pi\cdot 10^{-7} H/m) (3\cdot 10^8 m/s)}=127.5 W/m^2$

(g) $1.53\cdot 10^{-8} kg m/s$

For a surface that totally absorbs the wave, the rate at which momentum is transferred to the surface given by

$\frac{dp}{dt}=\frac{A}{c}$

where the <S> is the magnitude of the Poynting vector, given by

$=\frac{EB}{\mu_0}=\frac{(310 V/m)(1.03\cdot 10^{-6} T)}{4\pi \cdot 10^{-7}H/m}=254.2 W/m^2$

and where the surface is

A = 1.8 m^2

Substituting, we find

$\frac{dp}{dt}=\frac{(254.2 W/m^2)(1.8 m^2)}{3\cdot 10^8 m/s}=1.53\cdot 10^{-8} kg m/s$

(h) $8.47\cdot 10^{-7} N/m^2$

For a surface that totally absorbs the wave, the radiation pressure is given by

$p=\frac{}{c}$

where we have

$=254.2 W/m^2$

$c=3\cdot 10^8 m/s$

Substituting, we find

$p=\frac{254.2 W/m^2}{3\cdot 10^8 m/s}=8.47\cdot 10^{-7} N/m^2$

8 0

3 years ago

Help with questions 16 and 17! Please provide explanations too!

jok3333 [9.3K]

Please look at the attached awesome drawing.

Both answers are there.

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3 years ago

A size-5 soccer ball of diameter 22.6 cm and mass 426 g rolls up a hill without slipping, reaching a maximum height of 5.00 m ab

maria [59]

Answer:

W = 0.678 rad/s

Explanation:

Using the conservation of energy:

$E_i =E_f$

Roll up and hill without slipping is the sumatory of two energys, rotational and translational, so:

$\frac{1}{2}IW^2+ \frac{1}{2}mV^2 = mgh$

where I is the moment of inertia, W the angular velocity at the base of the hill, m the mass of the ball, V the velocity at the base of the hill, g the gravity and h the altitude.

First, we will find the moment of inertia as:

I = $\frac{2}{3}mR^2$

where m is the mass and R the radius, so:

I = $\frac{2}{3}(0.426kg)(11.3m)^2$

I = 36.26 Kg*m^2

Then, replacing values on the initial equation, we get:

$\frac{1}{2}(36.26)W^2+ \frac{1}{2}(0.426kg)V^2 = (0.426kg)(9.8)(5m)$

also we know that:

V =WR

so:

$\frac{1}{2}(36.26)W^2+ \frac{1}{2}(0.426kg)W^2R^2 = (0.426kg)(9.8)(5m)$

Finally, solving for W, we get:

$W^2(\frac{1}{2}(36.26)+ \frac{1}{2}(0.426kg)(11.3m)^2) = (0.426kg)(9.8)(5m)$

W = 0.678 rad/s

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3 years ago