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3 years ago

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The ease with which the charge distribution in a molecule can be distorted by an external electrical field is called the _______

___.
A) electronegativity
B) hydrogen bonding
C) polarizability
D) volatility
E) viscosity

1 answer:

Greeley [361]3 years ago

4 0

Answer:

C

Explanation:

The answer is polarization. In a normal molecule the positive and negative charges are distributed in a way that it makes a continuous distribution with no net force. but when you apply an external electric force, the opposite charges go to <em>the opposite ends of the molecule making them a polarized</em> molecule.

<em>This is called polarization.</em>

I hope it helps you.

Thank you.

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The work function of cesium is 1.96 eV. If radiation of wavelength 4.00 × 10^2 nm is incident on the surface, find the kinetic e

nata0808 [166]

I have a note here that might help you solve the problem on your own:

According to the law of conservation of energy

The energy of the photon is used in overcoming the workfunction of the metal, and the excess energy goes as the kinetic energy of the electron.

<span>Photon Energy = Work function of Cs + KE of electron
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I hope my guide has come to your help. God bless and have a nice day ahead!

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3 years ago

Se lanza un cohete en un ángulo de 53° sobre la horizontal con una rapidez inicial de 100 m/s. El cohete se mueve por

likoan [24]

Answer:

Explanation:

v = u + at

v₃ = 100 +30.0(3.00) = 190 m/s

s = vt + ½at²

y₃ = (100sin53)(3.00) + ½(30sin53)(3.00²) = 347.4 m

x₃ = (100cos53)(3.00) + ½(30cos53)(3.00²) = 261.8 m

a) v² = u² + uas

s = (v² - u²) / 2a

ymax = 347.4 + (0² - (190sin53)²) / (2(-9.80)) = 1,522 m

b) t₁ = 3.00 s

t₂ = (190sin53) / 9.80 = 15.5 s

t₃ = √(2(1522) / 9.80) = 17.6 s

t = 3.00 + 15.5 + 17.6 = 36.1 s

c) xmax = 261.8 + (190cos53)( 15.5 + 17.6) = 4,047 m

3 0

3 years ago

Suppose that the strings on a violin are stretched with the same tension and each has the same length between its two fixed ends

lesya692 [45]

Answer:

0.00384 kg/m

Explanation:

The fundamental frequency of string waves is given by

$f=\frac{1}{2L}\sqrt{\frac{F}{\mu}}$

For some tension (F) and length (L)

$f\propto\frac{1}{\mu}$

Fundamental frequency of G string

$f_G=196\ Hz$

Fundamental frequency of E string

$f_E=659.3\ Hz$

Linear mass density of E string is

$\mu_E=3.4\times 10^{-4}\ kg/m$

So,

$\frac{F_G}{F_E}=\sqrt{\frac{\mu_E}{\mu_G}}\\\Rightarrow \frac{F_G^2}{F_E^2}=\frac{\mu_E}{\mu_G}\\\Rightarrow \mu_G=3.4\times 10^{-4}\times \frac{659.3^2}{196^2}\\\Rightarrow \mu_G=0.00384\ kg/m$

The linear density of the G string is 0.00384 kg/m

4 0

3 years ago

What is maxwell’s theory about atoms?

Inga [223]

To put it another way, stationary states are the same as time-independent currents. Moreover, such currents are not sources of radiation, according to Maxwell's theory. The atom will remain in its natural state and will not collapse due to a shortage of energy.

4 0

3 years ago

Read 2 more answers

Es muy común que cuando se viaja hacia un río o lago se juegue "ranita", el cual consiste en lanzar una piedra horizontalmente h

shutvik [7]

Answer:

a) La piedra es lanzada desde una altura de 0,785 metros.

b) La piedra es lanzada con una velocidad inicial de 6,25 metros por segundo.

Explanation:

a) Dado que la piedra es lanzada horizontalmente, tenemos que la piedra experimenta un movimiento horizontal a velocidad constante y uno vertical uniformemente acelerado debido a la gravedad. La altura de la que fue lanzada la piedra se puede determinar mediante la siguiente ecuación cinemática:

$y = y_{o}+v_{o,y}\cdot t +\frac{1}{2}\cdot g\cdot t^{2}$ (1)

Donde:

$y$ - Altura final, medida en metros.

$y_{o}$ - Altura inicial, medida en metros.

$v_{o,y}$ - Componente vertical de la velocidad inicial, medida en metros por segundo.

$t$ - Tiempo, medido en segundos.

$g$ - Aceleración gravitacional, medida en metros por segundo cuadrado.

Si sabemos que $y = 0\,m$ , $v_{o,y} = 0\,\frac{m}{s}$ , $t = 0,4\,s$ y $g = -9,807\,\frac{m}{s^{2}}$ , entonces la altura inicial de la piedra es:

$y_{o} = y-v_{o,y}\cdot t -\frac{1}{2}\cdot g\cdot t^{2}$

$y_{o} = 0\,m-\left(0\,\frac{m}{s} \right)\cdot (0,4\,s)-\frac{1}{2}\cdot \left(-9,807\,\frac{m}{s^{2}} \right) \cdot (0,4\,s)^{2}$

$y_{o} = 0,785\,m$

La piedra es lanzada desde una altura de 0,785 metros.

b) Ahora, obtenemos el componente horizontal de la velocidad inicial a partir de la siguiente ecuación cinemática:

$v_{o,x} = \frac{x-x_{o}}{t}$ (2)

Donde:

$x_{o}$ , $x$ - Posiciones horizontales iniciales y finales, medidas en metros.

$t$ - Tiempo, medido en segundos.

Si tenemos que $x_{o} = 0\,m$ , $x = 2,5\,m$ y $t = 0,4\,s$ , entonces el componente horizontal de la velocidad inicial es:

$v_{o,x} = \frac{2,5\,m-0\,m}{0,4\,s}$

$v_{o,x} = 6,25\,\frac{m}{s}$

La piedra es lanzada con una velocidad inicial de 6,25 metros por segundo.

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3 years ago