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asambeis [7]
4 years ago
12

Before the development of quantum theory, Ernest Rutherford's experiments with gold atoms led him to propose the so-called Ruthe

rford Model of atomic structure. The basic idea is that the nucleus of the atom is a very dense concentration of positive charge, and that negatively charged electrons orbit the nucleus in much the same manner as planets orbit a star. His experiments appeared to show that the average radius of an electron orbit around the gold nucleus must be about 10−1010−10 m. Stable gold has 79 protons and 118 neutrons in its nucleus.
What is the strength of the nucleus' electric field at the orbital radius of the electrons?
What is the kinetic energy of an electron in a circular orbit around the gold nucleus?
Physics
1 answer:
NNADVOKAT [17]4 years ago
3 0

Answer:

a)    F = -1.82 10⁻¹⁵ N,  b) K = 9.1 10⁻¹⁶ J

Explanation:

a) To calculate the force between the nucleus and the electrons, let's use the Coulomb equation

           F = k q Q / r²

as the nucleus occupies a very small volume compared to electrons, we can suppose it as punctual

let's calculate

          F = 9 10⁹ (-1.6 10⁻¹⁹) (79 1.6 10⁻¹⁹) / (10⁻¹⁰)²

          F = -1.82 10⁻¹⁵ N

b) they ask us for kinetic energy

let's use Newton's second law

         F = m a

acceleration is centripetal

         a = v² / r

we substitute

         F = m v² / r

         v = √ (F r / m)

         v = √ (1.82 10⁻¹⁵ 10⁻¹⁰ / 9.1 10⁻³¹)

         v = √ (0.2 10⁻¹⁶)

         v = 0.447 10⁸ m / s

kinetic energy is

          K = ½ m v²

          K = ½ 9.1 10⁻³¹ (0.447 10⁸)²

          K = 0.91 10⁻¹⁵ J

          K = 9.1 10⁻¹⁶ J

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Stand on a bathroom scale on a level floor, and the reading on the scale shows the gravitational force on you, mg. If the floor
kari74 [83]

Answer:

<em>The angle introduces an error on the mesure of the weight</em>

Explanation:

<u><em>Weight and Normal Forces</em></u>

When an object is resting on a horizontal surface, its weight is directed downwards and the normal force has the same magnitude and opposite direction, i.e. directed upwards. When some angle α exists between the surface and the horizontal plane, the scale keeps 'feeling' the Normal force, but it's not equal to the weight anymore, but to the perpendicular component of the weight to the surface where the scale is placed. It can be found that the component of the weight is m.g.cosα

If, for example \alpha=10^o, the real measure is

m.g.cos 10^o=0.985(mg)

That is, 98.5% of the correct weight. So it's important to measure weight on horizontal surfaces

7 0
4 years ago
If the direction changes as a motorcycle goes around the curve, what else must be changing?A. AccelerationB. ForceC. SpeedD. Vel
pogonyaev

In physics, circular motion is a movement of an object along the circumference of a circle or rotation along a circular path. It can be uniform, with constant angular rate of rotation and constant speed, or non-uniform with a changing rate of rotation. Since the object's velocity vector is constantly changing direction, the moving object is undergoing acceleration by a centripetal force in the direction of the center of rotation. Without this acceleration, the object would move in a straight line.

In this sense, the acceleration is always changing due to centripetal acceleration.

Answer:

A. Acceleration

6 0
1 year ago
A cyclist is riding a bicycle at a speed of 22 mph on a horizontal road. The distance between the axles is 42 in., and the mass
stealth61 [152]

Answer:

The shortest distance is  S = 24.86 ft

Explanation:

The free body diagram of this question is shown on the first uploaded image

From the question we are told that

   The speed of the bicycle is v_b = 22\ mph = 22 * \frac{5280}{3600}   =  32.26 ft/s

     The distance between the axial is  d = 42 \ in

The mass center of the cyclist and the bicycle is m_c = 26 \ in  behind the front axle

The mass center of the cyclist and the bicycle is m_h = 40 \ in above the ground

   For the bicycle not to be thrown over the

     Momentum about the back wheel must be zero so

                \sum _B = 0

=>             mg (26) = ma(40)

=>             a = \frac{26}{40} g

Here  g = 32.2 ft/s^2

     So     a =  \frac{26}{40} (32.2)

             a =  20.93 ft/s^2

Apply the equation of motion to this motion we have

       v^2 = u^2 + 2as

 Where  u = 32.26 ft /s

             and v = 0 since the bicycle is coming to a stop

        v^2 = (32.26)^2 - 2(20.93) S

=>      S = 24.86 ft        

                 

5 0
3 years ago
Suppose you observed the equation for a traveling wave to be y(x, t) = A cos(kx − ????t), where its amplitude of oscillations wa
OLga [1]

Answer:

<h2>15m/s</h2>

Explanation:

The equation for a traveling wave as expressed as y(x, t) = A cos(kx − \omegat) where An is the amplitude f oscillation, \omega is the angular velocity and x is the horizontal displacement and y is the vertical displacement.

From the formula; k =\frac{2\pi x}{\lambda} \ and \ \omega = 2 \pi f where;

\lambda \ is\ the \ wavelength \ and\ f \ is\ the\ frequency

Before we can get the transverse speed, we need to get the frequency and the wavelength.

frequency = 1/period

Given period = 2/15 s

Frequency = \frac{1}{(2/15)}

frequency = 1 * 15/2

frequency f = 15/2 Hertz

Given wavelength \lambda = 2m

Transverse speed v = f \lambda

v = 15/2 * 2\\\\v = 30/2\\\\v = 15m/s

Hence, the transverse speed at that point is  15m/s

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3 years ago
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3 years ago
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