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Gre4nikov [31]
3 years ago
11

The deflection of alpha particles in rutherford’s gold foil experiments resulted in what change to the atomic model? the additio

n of energy levels and orbitals to the electron cloud the addition of a small, dense nucleus to the center of the atom the addition of wave properties to the characteristics of an electron the addition of neutrons to the nucleus of the atom
Physics
2 answers:
skelet666 [1.2K]3 years ago
5 0
The answer is: The addition of a small dense nucleus at the center of the atom.
pogonyaev3 years ago
4 0

Answer:

Addition of a small, dense nucleus to the center of the atom

Explanation:

Most alpha particles were observed to pass straight through the gold foil, which implied that atoms are composed of large amounts of open space. Some alpha particles were deflected slightly, suggesting interactions with other positively charged particles within the atom.

Still other alpha particles were scattered at large angles, while a very few even bounced back toward the source. Only a positively charged and relatively heavy target particle, such as the proposed nucleus, could account for such strong repulsion.

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Red light strikes a metal surface and electrons are ejected. If violet light is now used with a 10% greater intensity, what will
Juli2301 [7.4K]

Answer:

correct answer is C

Explanation:

The photoelectric effect was correctly described by Einstein, where he assumes that the light ray is formed by photons that are articulated and behaves like an elastic shock, the energy of this particular is described by the Planck equation.

             K = h f + Ф

where k is the kinetic energy of the electrons, f the frequency of the photons and Ф the work function of the material.

In this experiment, red light removes electrons, it is assumed that each photon spreads an electron if we have another light with more energy and 10% more intense, that is, with 10% more shapes and each arcane an electron the number of electrons removed of; material is increased by 10%.

The change in wavelength and consequently the frequency

            c = λ f

            f = c /λ

therefore, the wavelength of the voilet λ = 400 num has a higher frequency and therefore more energy, so that the turned-on turns have more kinetic energy.

With these approaches we examine the final answers where the correct answer is C

3 0
3 years ago
A 7.0kg object rests on a horizontal frictionless surface. What is the magnitude of the horizontal
sukhopar [10]

Answer:

16.1 N

Explanation:

From the question,

F = ma.............................. Equation 1

Where F = horizontal force, m = mass of the object, a = acceleration .

Given: m = 7.0 kg, a = 2.3 m/s²

Substitute this values into equation 1

F = (7.0×2.3)

F = 16.1 N.

Hence the magnitude of the horizontal force is 16.1 N

6 0
3 years ago
If we have less power, we most likely have
katrin [286]
<span>the same amount of work being done over a longer period of time.</span>
3 0
3 years ago
Read 2 more answers
When an automobile moves with constant speed down a highway, most of the power developed by the engine is used to compensate for
sveta [45]

Answer:

F = -4567.40 N

Explanation:

Given that,

The power developed by the engine, P = 196 hp

1 hp = 746 W

196 hp = 146157 W

Speed of the car, v = 32 m/s

Let F is the total friction force acting on the car. The product of force and velocity is called the power developed by the engine. It is given by :

P=-F\times v

F=\dfrac{-P}{v}

F=\dfrac{-146157\ W}{32\ m/s}

F = -4567.40 N

So, the total frictional force acting on the car is 4567.40 N. Hence, this is the required solution.

7 0
3 years ago
Please help im being timed on this and need to get my grade up
nadya68 [22]

Answer:

a I think hope this helps

8 0
3 years ago
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