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3 years ago

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An electron and a proton have the same kinetic energy upon entering a region of constant magnetic field and their velocity vecto

rs are perpendicular to the magnetic field. Suppose the magnetic field is strong enough to allow the particles to circle in the field. Note: you'll need to look up the masses for an electron and proton. 1) What is the ratio of the radii of their circular paths rp/re?

1 answer:

kupik [55]3 years ago

4 0

Answer: rp/re= me/mp= 544 * 10^-6.

Explanation: To calculate this problem we have to consider the circular movement by the electron and proton inside a magnetic field.

Then the dynamic equation for the circular movement is given by:

Fcentripetal= m*ω^2.r

q*v*B=m*ω^2.r

we write this for each particle then we have the following:

q*v*B=me* ω^2*re

q*v*B=mp* ω^2*rp

rp/re=me/mp=9.1*10^-31/1.67*10^-27=544*10^-6

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Consider a string of total length L, made up of three segments of equal length. The mass per unit length of the first segment is

zzz [600]

Answer:

Explanation:

Length of each segment is $\frac{L}{3}$

Speed of wave in first segment is $v_1=\sqrt{\frac{T_s}{\mu}}$

Speed of wave in second segment is $v_2=\sqrt{\frac{T_s}{2\mu}}$

Speed of wave in third segment is $v_3=\sqrt{\frac{T_s}{\frac{\mu}{4}}}=\sqrt{\frac{4T_s}{\mu}}$

Now time for the transverse wave to propagate is

$t=t_1 + t_2 + t_3\\=\frac{(\frac{L}{3})}{v_1}+\frac{(\frac{L}{3})}{v_2} + \frac{(\frac{L}{3})}{v_3}\\\\=(\frac{L}{3})(\frac{1}{\sqrt{\frac{T_s}{\mu}}} + \frac{1}{\sqrt{\frac{T_s}{2\mu}}} + \frac{1}{\sqrt{\frac{4T_s}{\mu}}})$

simplifying we get

$t=(\frac{3+2\sqrt{2}}{6})L\sqrt{\frac{\mu}{T_s}}$

3 0

3 years ago

Determine the average value of the translational kinetic energy of the molecules of an ideal gas at (a) 27.8°C and (b) 143°C. Wh

Alinara [238K]

Answer:

a) $k_{avg}=6.22\times 10^{-21}$

b) $k_{avg}=8.61\times 10^{-21}$

c) $k_{mol}=3.74\times 10^{3}J/mol$

d) $k_{mol}=5.1\times 10^{3}J/mol$

Explanation:

Average translation kinetic energy ( $k_{avg}$ ) is given as

$k_{avg}=\frac{3}{2}\times kT$ ....................(1)

where,

k = Boltzmann's constant ; 1.38 × 10⁻²³ J/K

T = Temperature in kelvin

a) at T = 27.8° C

or

T = 27.8 + 273 = 300.8 K

substituting the value of temperature in the equation (1)

we have

$k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 300.8$

$k_{avg}=6.22\times 10^{-21}J$

b) at T = 143° C

or

T = 143 + 273 = 416 K

substituting the value of temperature in the equation (1)

we have

$k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 416$

$k_{avg}=8.61\times 10^{-21}J$

c ) The translational kinetic energy per mole of an ideal gas is given as:

$k_{mol}=A_{v}\times k_{avg}$

here $A_{v}$ = Avagadro's number; ( 6.02×10²³ )

now at T = 27.8° C

$k_{mol}=6.02\times 10^{23}\times 6.22\times 10^{-21}$

$k_{mol}=3.74\times 10^{3}J/mol$

d) now at T = 143° C

$k_{mol}=6.02\times 10^{23}\times 8.61\times 10^{-21}$

$k_{mol}=5.1\times 10^{3}J/mol$

8 0

3 years ago

N a scientific investigation, what is the name for a prediction that can be tested?

ELEN [110]

A hypothesis is an educated prediction that can be tested.

8 0

3 years ago

If you were to be drawn into a black hole, what would happen? To the black hole, not to you.

love history [14]

Answer:

It grows

Explanation:

The blacks holes will absorb

Me hoizontally stretching me like a noodle by the spaghtification process,thus growing bigger.

7 0

3 years ago

Calcula el valor de la velocidad de las ondas sonoras en el agua sabiendo que su

dybincka [34]

La velocidad de las ondas sonoras es aproximadamente 1469,694 metros por segundo.
La longitud de onda de las ondas sonoras es 1,470 metros.

1) Inicialmente, debemos determinar la velocidad de las ondas sonoras a través del agua ( $v$ ), en metros por segundo:

$v = \sqrt{\frac{K}{\rho} }$ (1)

Donde:

$K$ - Módulo de compresibilidad, en newtons por metro cuadrado.
$\rho$ - Densidad del agua, en kilogramos por metro cúbico.

Si sabemos que $\rho = 1\times 10^{3}\,\frac{kg}{m^{3}}$ y $K = 2,16\times 10^{9}\,\frac{N}{m^{2}}$ , entonces la velocidad de las ondas sonoras es:

$v = \sqrt{\frac{2,16\times 10^{9}\,\frac{N}{m^{2}}}{1\times 10^{3}\,\frac{kg}{m^{3}} } }$

$v\approx 1469,694\,\frac{m}{s}$

La velocidad de las ondas sonoras es aproximadamente 1469,694 metros por segundo.

2) Luego, determinamos la longitud de onda ( $\lambda$ ), en metros, mediante la siguiente fórmula:

$\lambda = \frac{v}{f}$ (2)

Donde $f$ es la frecuencia de las ondas sonoras, en hertz.

Si sabemos que $v\approx 1469,694\,\frac{m}{s}$ y $f = 1000\,hz$ , entonces la longitud de onda de las ondas sonoras es:

$\lambda = \frac{1469,694\,\frac{m}{s} }{1000\,hz}$

$\lambda = 1,470\,m$

La longitud de onda de las ondas sonoras es 1,470 metros.

Para aprender más sobre las ondas sonoras, invitamos a ver esta pregunta verificada: brainly.com/question/1070238

6 0

2 years ago