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cupoosta [38]
3 years ago
10

Why has the atomic model changed over time

Physics
1 answer:
inysia [295]3 years ago
7 0
                                   

The first model of the atom was developed by JJ Thomson in 1904, who thought that atoms were composed purely of negatively charged electrons. This model was known as the 'plum pudding' model.

This theory was then disproved by Ernest Rutherford and the gold foil experiment in 1911, where Rutherford shot alpha particles at gold foil, and noticed that some went through and some bounced back, implying the existence of a positive nucleus.

In 1913, Niels Bohr proposed a model of the atom where the electrons were contained within quantized shells that orbited the nucleus. This was because it was impossible for the cloud of negative electrons proposed by Rutherford to exist, as the negative electrons would be drawn to the positive nucleus, and the atom would collapse in on itself.

In 1926, the Austrian physicist Erwin Schrödinger created a quantum mechanical model of the atom by combining the equations for the behavior of waves with the de Broglie equation to generate a mathematical model for the distribution of electrons in an atom.

However the model used today is closest to the Bohr model of the atom, using the quantized shells to contain the electrons.

For more info:

http://chemistry.about.com/od/chemistryglossary/a/debroglieeqdef.htm

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A projector is placed on the ground 22 ft. away from a projector screen. A 5.2 ft. tall person is walking toward the screen at a
Stella [2.4K]

Answer:

y = 67.6 feet,   y = 114.4/ (22 - 3t)

Explanation:

For this exercise let's use that light travels in a straight line and some trigonometric relationships, the symbols are in the attached diagram

Large triangle Projector up to the screen

         tan θ = y / L

For the small triangle. Projector up to the person

         tan θ = y₀ / (L-d)

The angle is the same, so we equate the two equations

         y₀ / (L -d) = y / L

         y = y₀  L / (L-d)

The distance from the screen (d), we look for it with kinematics

         v = d / t

        d = v t

we replace

         y = y₀ L / (L - v t)

         y = 5.2 22 / (22 - 3 t)

         y = 114.4 (22 - 3t)⁻¹

This is the equation of the shadow height change as a function of time

For the suggested distance the shadow has a height of

           y = 114.4 / (22-13)

           y = 67.6 feet

7 0
3 years ago
A car is traveling at 75 m/s. 50 seconds later it is traveling at 25 m/s. What is the car’s acceleration?
scoray [572]

Answer:

a = -1 m/s^2

Explanation:

Vi = 75 m/s

Vf = 25 m/s

t = 50 s

Plug those values into the following equation:

Vf = Vi + at

25 = 75 + 50a

---> a = -1 m/s^2

5 0
3 years ago
Question 4 (18 marks) (a) During a Physics Lab experiment, 1 st year SFY students analyzed the behavior of capacitors by connect
Nataly_w [17]

Answer:

1.) 274.5v

2.) 206.8v

Explanation:

1.) Given that In one part of the lab activities, students connected a 2.50 µF capacitor to a 746 V power source, whilst connected a second 6.80 µF capacitor to a 562 V source.

The potential difference and charge across EACH capacitor will be

V = Voe

Where Vo = initial voltage

e = natural logarithm = 2.718

For the first capacitor 2.50 µF,

V = Vo × 2.718

746 = Vo × 2.718

Vo = 746/2.718

Vo = 274.5v

To calculate the charge, use the below formula.

Q = CV

Q = 2.5 × 10^-6 × 274.5

Q = 6.86 × 10^-4 C

For the second capacitor 6.80 µF 

V = Voe

562 = Vo × 2.718

Vo = 562/2.718

Vo = 206.77v

The charge on it will be

Q = CV

Q = 6.8 × 10^-6 × 206.77

Q = 1.41 × 10^-3 C

B.) Using the formula V = Voe again

165 = Vo × 2.718

Vo = 165 /2.718

Vo = 60.71v

Q = C × 60.71

Q = C

4 0
3 years ago
The earth rotates about its pole once every 24 hrs. The distance from the pole to a location on the Earth is 35* north latitude
Oksana_A [137]
The angular velocity, ω= 
2π/t; t = 24 hrs = 24 x 3600 seconds = 86400 s
ω = 7.27 x 10⁻⁵
v = ωr
= 7.27 x 10⁻⁵ x 3242.8 x 1.6 x 1000 (converting miles to meters)
= 377.2 m/s
4 0
3 years ago
PLEASE HELP The graph shows the amplitude of a passing wave over time in seconds (s). What is the approximate frequency of the w
d1i1m1o1n [39]

Answer:

F = 0.3\ Hz

Explanation:

Given

See attachment for the graph

Required

Determine the frequency

Frequency (F) is calculated as:

F = \frac{1}{T}

Where

T = Time to complete a period

From the attachment, the wave complete a cycle or period in 3 seconds..

So:

F = \frac{1}{3s}

F = 0.333\ Hz

F = 0.3\ Hz --- Approximated

7 0
3 years ago
Read 2 more answers
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