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Wewaii [24]
3 years ago
7

Oppositely charged objects attract each other. This attraction holds electrons in atoms and holds atoms to one another in many c

ompounds. However, Ernest Rutherford’s model of the atom failed to explain why electrons were not pulled into the atomic nucleus by this attraction. What change to the atomic model helped solve the problem seen in Rutherford’s model?
Physics
2 answers:
GalinKa [24]3 years ago
6 0

Answer and Explanation:

This can be explained as in Rutherford's model of atom the electrons orbits the nucleus which means that they will travel around the nucleus with some velocity and hence radiate electromagnetic waves which results in the loss of energy due to which the electron keeps coming closer and eventually falls into the nucleus.

But Bohr came up with a better explanation as according to the Bohr's atomic model, electrons stay fixed in orbit with certain energy in different shells around the nucleus and can only jump from an energy level to another if that specific amount of energy is supplied to it.

This model is based on the quantization of energy thus giving an explanation why electrons do not fall into the nucleus of an atom.

strojnjashka [21]3 years ago
6 0

Answer:

A) Bohr’s work with atomic spectra led him to say that the electrons were limited to existing in certain energy levels, like standing on the rungs of a ladder.

Explanation:

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A volcano erupts spewing ash into the air and sending lava flowing down the side of the mountain. Looking at the image explain h
Anit [1.1K]

Answer:

A lateral eruptions or lateral blast is a volcanic eruption which is directed laterally from a volcano rather than upwards from the summit. Lateral eruptions are caused by the outward expansion of flanks due to rising magma. Breaking occurs at the flanks of volcanoes making it easier for magma to flow outward.

Explanation:

7 0
3 years ago
An electron is released from rest at a distance of 6.00 cm from a proton. If the proton is held in place, how fast will the elec
lana66690 [7]

Answer:

91.87 m/s

Explanation:

<u>Given:</u>

  • x = initial distance of the electron from the proton = 6 cm = 0.06 m
  • y = initial distance of the electron from the proton = 3 cm = 0.03 m
  • u = initial velocity of the electron = 0 m/s

<u>Assume:</u>

  • m = mass of an electron = 9.1\times 10^{-31}\ kg
  • v = final velocity of the electron
  • e = magnitude of charge on an electron = 1.6\times 10^{-19}\ C
  • p = magnitude of charge on a proton = 1.6\times 10^{-19}\ C

We know that only only electric field due to proton causes to move from a distance of 6 cm from proton to 3 cm distance from it. This means the electric force force does work on the electron to move it from one initial position to the final position which is equal to the change in potential energy of the electron due to proton.

Now, according to the work-energy theorem, the total work done by the electric force on the electron due to proton is equal to the kinetic energy change in it.

\therefore \textrm{Kinetic energy change}= \textrm{Change in potential energy}\\\Rightarrow \dfrac{1}{2}m(v^2-u^2)= \dfrac{kpe}{y}-\dfrac{kpe}{x}\\\Rightarrow \dfrac{1}{2}m(v^2-(0)^2)= \dfrac{kpe}{0.03}-\dfrac{kpe}{0.06}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{3}-\dfrac{100kpe}{6}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{6}\\

\Rightarrow v^2= \dfrac{100kpe\times 2}{6m}\\\Rightarrow v^2= \dfrac{100kpe}{3m}\\\Rightarrow v^2= \dfrac{100\times 9\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{3\times 9.1\times 10^{-31}}\\\Rightarrow v^2=8.44\times 10^3\\\Rightarrow v=91.87\ m/s\\

Hence, when the electron is at a distance of c cm from the proton, it moves with a velocity of 91.87 m/s.

8 0
3 years ago
A car accelerate from a stop to 27 m/sec in 5 seconds. What was the car's acceleration?
Furkat [3]
The acceleration should 5.4 m/s^2
4 0
3 years ago
The mean time between collisions for electrons in room temperature copper is 2.5 x 10-14 s. What is the electron current in a 2
Darya [45]

Answer:

1.87 A

Explanation:

τ = mean time between collisions for electrons = 2.5 x 10⁻¹⁴ s

d = diameter of copper wire = 2 mm = 2 x 10⁻³ m

Area of cross-section of copper wire is given as

A = (0.25) πd²

A = (0.25) (3.14) (2 x 10⁻³)²

A = 3.14 x 10⁻⁶ m²

E = magnitude of electric field = 0.01 V/m

e = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

m = mass of electron = 9.1 x 10⁻³¹ kg

n = number density of free electrons in copper = 8.47 x 10²² cm⁻³ = 8.47 x 10²⁸ m⁻³

i = magnitude of current

magnitude of current is given as

i = \frac{Ane^{2}\tau E}{m}

i = \frac{(3.14\times 10^{-6})(8.47\times 10^{28})(1.6\times 10^{-19})^{2}(2.5\times 10^{-14}) (0.01)}{(9.1\times 10^{-31})}

i  = 1.87 A

4 0
3 years ago
The law of reflection states that if the angle of incidence is 19 degrees, the angle of reflection is ___ degrees.
Wewaii [24]
using the law of refraction, the incidence is equal to the reflection, but not refraction
4 0
3 years ago
Read 2 more answers
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