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beks73 [17]
3 years ago
5

During a famous experiment in 1919, Ernest Rutherford shot doubly ionized helium nuclei (also known as alpha particles) at a gol

d foil. He discovered that virtually all of the mass of an atom resides in an extremely compact nucleus. Suppose that during such an experiment, an alpha particle far from the foil has a kinetic energy of 6.0 MeV. If the alpha particle is aimed directly at the gold nucleus, and the only force acting on it is the electric force of repulsion exerted on it by the gold nucleus, how close will it approach the gold nucleus before turning back
Physics
2 answers:
Delvig [45]3 years ago
5 0

Answer:

The answer to the question is;

The alpha particle will approach the gold nucleus up to 3.79 × 10⁻¹⁴ m before turning back.

Explanation:

To solve the question, we note that

There are  79 electrons the gold tom and 2 electrons in the alpha particle which is the helium nuclear

We use the energy conservation principle to obtain

W_{net} = \int\limits^{r_{min}}_{inf} {F_e} \, dr = Δ KE

Where

F_e = Coulomb force

r = Distance of the alpha particle from the gold foil

we get from F = k(Q₁·Q₂)/r²

W_{net} = \int\limits^{r_{min}}_{inf} {F_e} \, dr =(-1)× \int\limits^{r_{min}}_{inf} {\frac{k(2e)(79e)}{r^2} } \, dr

= -158·k·e² × \int\limits^{r_{min}}_{inf} {\frac{dr}{r^2} } \,⇒ -158·k·e²[\frac{1}{r} ]^{r_{min}} _{inf} = \frac{-158*k*e^2}{r_{min}} = -KE₁

or r_{min} = (158·k·e²)/KE₁

Where Δ KE = - KE₁ as KE₂ = 0, particle at rest

Solving gives

r_{min} =\frac{158*(8.988*10^9\frac{Nm^2}{C^2} )(1.602*10^{-19}C)^2}{6Mev*\frac{1.602*10^{-19}J}{eV} *1000000}     = 3.79 × 10⁻¹⁴ m

NISA [10]3 years ago
4 0

Answer:

r = 3.79 10⁻¹⁵ m

Explanation:

For this exercise we can use energy conservation

Starting point. Far from the fixed core

       Em₀ = K

Final point

        Em_{f} = U = k q1 q2 / r

           

      Em₀ =Em_{f}

       K = k q₁ q₂ / r

       r = k q₂ q₂ / K

Data

     K = 6 Mev = 6 106 eV (1.6 10⁻¹⁹ J / 1 eV) = 9.6 10⁻¹³ J

    Helium

    q₁ = 2 e = 2 1.6 10⁻¹⁹ C = 3.2 10⁻¹⁹ C

     Gold

    q2 = 79 e = 79 1.6 10⁻¹⁹ = 126.4 10⁻¹⁹ C

Let's calculate

      r = 8.99 10⁸ 3.2 10⁻¹⁹ 126.4 10⁻¹⁹ / 9.6 10⁻¹³

      r = 3.79 10⁻¹⁵ m

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Answer:

Explanation:

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(a) yes, there will be tangential velocity which is given by

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An airliner arrives at the terminal, and its engines are shut off. The rotor of one of the engines has an initial clockwise angu
Ilia_Sergeevich [38]

(a) 1200 rad/s

The angular acceleration of the rotor is given by:

\alpha = \frac{\omega_f - \omega_i}{t}

where we have

\alpha = -80.0 rad/s^2 is the angular acceleration (negative since the rotor is slowing down)

\omega_f is the final angular speed

\omega_i = 2000 rad/s is the initial angular speed

t = 10.0 s is the time interval

Solving for \omega_f, we find the final angular speed after 10.0 s:

\omega_f = \omega_i + \alpha t = 2000 rad/s + (-80.0 rad/s^2)(10.0 s)=1200 rad/s

(b) 25 s

We can calculate the time needed for the rotor to come to rest, by using again the same formula:

\alpha = \frac{\omega_f - \omega_i}{t}

If we re-arrange it for t, we get:

t = \frac{\omega_f - \omega_i}{\alpha}

where here we have

\omega_i = 2000 rad/s is the initial angular speed

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\alpha = -80.0 rad/s^2 is the angular acceleration

Solving the equation,

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6 0
3 years ago
Calculate the wavelength of each frequency of electromagnetic radiation: a. 100.2 MHz (typical frequency for FM radio broadcasti
Natalka [10]

Answer:

a). 100.2 MHz (typical frequency for FM radio broadcasting)

The wavelength of a frequency of 100.2 Mhz is 2.99m.

b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)

The wavelength of a frequency of 1070 khz is 280.3 m.

c. 835.6 MHz (common frequency used for cell phone communication)

The wavelength of a frequency of 835.6 Mhz is 0.35m.

Explanation:

The wavelength can be determined by the following equation:

c = \lambda \cdot \nu  (1)

Where c is the speed of light, \lambda is the wavelength and \nu is the frequency.  

Notice that since it is electromagnetic radiation, equation 1 can be used. Remember that light propagates in the form of an electromagnetic wave.

<em>a). 100.2 MHz (typical frequency for FM radio broadcasting)</em>

Then, \lambda can be isolated from equation 1:

\lambda = \frac{c}{\nu} (2)

since the value of c is 3x10^{8}m/s. It is necessary to express the frequency in units of hertz.

\nu = 100.2 MHz . \frac{1x10^{6}Hz}{1MHz} ⇒ 100200000Hz

But 1Hz = s^{-1}

\nu = 100200000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{100200000s^{-1}}

\lambda = 2.99 m

Hence, the wavelength of a frequency of 100.2 Mhz is 2.99m.

<em>b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)</em>

<em> </em>

\nu = 1070kHz . \frac{1000Hz}{1kHz} ⇒ 1070000Hz

But  1Hz = s^{-1}

\nu = 1070000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{1070000s^{-1}}

\lambda = 280.3 m

Hence, the wavelength of a frequency of 1070 khz is 280.3 m.

<em>c. 835.6 MHz (common frequency used for cell phone communication) </em>

\nu = 835.6MHz . \frac{1x10^{6}Hz}{1MHz} ⇒ 835600000Hz

But  1Hz = s^{-1}

\nu = 835600000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{835600000s^{-1}}

\lambda = 0.35 m

Hence, the wavelength of a frequency of 835.6 Mhz is 0.35m.

6 0
3 years ago
6. A Cadillac Escalade has a mass of 2 569.6 kg, if it accelerates at 4.65m/S<br> force on the car?
SVETLANKA909090 [29]

Answer:

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*Multiply the mass and the acceleration to find the force

Explanation:

5 0
3 years ago
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