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dybincka [34]
4 years ago
7

Match the scientists with the correct atomic models.

Physics
1 answer:
bonufazy [111]4 years ago
4 0

Answer:

Explanation will be answer.

Explanation:

Democritus thought if he cut something in half over and over again, it would stop at some point. He said that this last piece of material couldn't be cut smaller. Democritus called these little pieces of atoms of matter, which means "indivisible". He believed that atoms last forever, never change, and could not be destroyed. Democritus believed that there was nothing between atoms and that everything around us could be explained if we could understand how atoms work.

The plum pudding model is one of many historical and scientific models of the atom. The model was first proposed by JJ Thomson in 1904, shortly after the discovery of the electron but before the discovery of the atomic nucleus, and tried to explain two properties of the atoms known at the time: that the electrons are negatively charged particles and the atoms are not. You have a net electric charge. The plum pudding model has electrons that are surrounded by a positively charged volume, like negatively charged "plums" that are embedded in a positively charged pudding.

In 1920 Ernest Rutherford imagined the possible existence of the neutron. In particular, Rutherford examined the inequality between the atomic number of an atom and its atomic mass. His explanation for this was the existence of a neutrally charged particle in the atomic nucleus. He viewed the neutron as a double neutral conductor made up of an electron orbiting a proton.

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A force gives a 5.0 kg object an acceleration of 2.0 m/s 2. The same force would give a 20 kg object an acceleration of _____. 0
Oksi-84 [34.3K]

m = 5 kg

a = 2 m/s²

to find the force that accelerates the 4 kg object @ 2 m/s²

F = ma = 5 kg x 2 m/s² = 10 N

To find what acceleration 10 N would give a 20 kg object

a = F/m = 10 N/20 kg = 0.5 m/s

6 0
3 years ago
Read 2 more answers
A plastic rod of length d = 1.5 m lies along the x-axis with its midpoint at the origin. The rod carries a uniform linear charge
Serga [27]

Answer:

Explanation:

Let the plastic rod extends from - L to + L .

consider a small length of dx on the rod on the positive x axis at distance x . charge on it =  λ dx where  λ is linear charge density .

It will create a field at point P on y -axis . Distance of point P

= √ x² + .15²

electric field at P due to small charged length

dE = k λ dx x  / (x² + .15² )

Its component along Y - axis

= dE cosθ where θ is angle between direction of field dE and y axis

= dE x .15 / √ x² + .15²

=  k λ dx  .15 / (x² + .15² )³/²

If we consider the same strip along the x axis at the same position  on negative x axis , same result will be found . It is to be noted that the component of field in perpendicular to y axis will cancel out each other . Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L

E = ∫  k λ  .15  / (x² + .15² )³/² dx

=  k λ  x L / .15 √( L² / 4 + .15² )

6 0
4 years ago
What graph shape is this what does the snap tell you
vekshin1
The picture is hard to see but if you still need help message me
7 0
3 years ago
What is the change in internal energy if 60 J of heat is released from a system and 30 J of work is done on the system? Use U =
k0ka [10]

The change in internal energy of the system is +30 J

Explanation:

We can solve this problem by using the first law of thermodynamics, which states that the change in internal energy of a system is given by the equation:

\Delta U = Q -W

where

\Delta U is the change in internal energy

Q is the heat absorbed by the system (positive if it is absorbed, negative if it is released)

W is the work done by the system (positive if it is done by the system, negative if it is done by the surroundings on the system)

Therefore, in this problem, we have

Q=-60 J (heat released by the system)

W=-30 J (work done on the system)

Therefore, the change in internal energy is

\Delta U = -60 - (-30) = +30 J

Learn more about thermodynamics:

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#LearnwithBrainly

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