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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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Pressure in a liquid increases with its ..........​
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Answer: Pressure increases as the depth increases

Explanation: The pressure in a liquid is due to the weight of the column of water above. Since the particles in a liquid are tightly packed, this pressure acts in all directions.

6 0
3 years ago
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A) It is very dry at the equator because it is so hot.
bixtya [17]

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6 0
3 years ago
The _________ ______ has the final say in both criminal and civil cases.
vagabundo [1.1K]
Judge/court has the final say
7 0
3 years ago
We can calculate the force that the atmospheric pressure produces on a surface. Consider a living room that has a 4.0m×5.0m floo
qaws [65]

Answer:

Force, F=2.02\times 10^6\ N

Explanation:

It is given that,

Length of the room, l = 4 m

breadth of the room, b = 5 m

Height of the room, h = 3 m

Atmospheric pressure, P=1\ atm=101325\ Pa

We know that the force acting per unit area is called pressure exerted. Its formula is given by :

P=\dfrac{F}{A}

F=P\times l\times b

F=101325\times 4\times 5

F=2.02\times 10^6\ N

So, the total force on the floor due to the air above the surface is 2.02\times 10^6\ N. Hence, this is the required solution.

4 0
3 years ago
If the earth's magnetic field has strength 0.50 gauss and makes an angle of 20.0 degrees with the garage floor, calculate the ch
lys-0071 [83]

Answer:

ΔΦ = -3.39*10^-6

Explanation:

Given:-

- The given magnetic field strength B = 0.50 gauss

- The angle between earth magnetic field and garage floor ∅ = 20 °

- The loop is rotated by 90 degree.

- The radius of the coil r = 19 cm

Find:

calculate the change in the magnetic flux δφb, in wb, through one of the loops of the coil during the rotation.

Solution:

- The change on flux ΔΦ occurs due to change in angle θ of earth's magnetic field B and the normal to circular coil.

- The strength of magnetic field B and the are of the loop A remains constant. So we have:

                         Φ = B*A*cos(θ)

                         ΔΦ = B*A*( cos(θ_1) - cos(θ_2) )

- The initial angle θ_1 between the normal to the coil and B was:

                         θ_1  = 90° -  ∅

                         θ_1  = 90° -  20° = 70°

The angle θ_2 after rotation between the normal to the coil and B was:

                         θ_2  =  ∅

                         θ_2  = 20°

- Hence, the change in flux can be calculated:

                        ΔΦ = 0.5*10^-4*π*0.19*( cos(70) - cos(20) )

                        ΔΦ = -3.39*10^-6

                       

8 0
3 years ago
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