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Masteriza [31]
2 years ago
8

The charged particles in the beams that Thomson studied came from atoms. As these particles moved away from their original atoms

, they formed a visible beam. The current model of the atom includes protons, neutrons, and electrons.
Physics
2 answers:
iris [78.8K]2 years ago
6 0

The current model of atom that includes protons, neutrons, and electrons is the model of Bohr. In his atomic model, the proton and neutrons are inside the nucleus and that the electrons orbit around the dense nucleus forming a visible beam.

aalyn [17]2 years ago
5 0

The question to the above information is;

What is the best use of an atomic model to explain the charge of the particles in Thomson's beams?

Answer;

An atom's smaller negative particles are at a distance from the central positive particles, so the negative particles are easier to remove.

Explanation;

-Atoms are comprised of a nucleus consisting of protons (red) and neutrons (blue). The number of orbiting electrons is the same as the number of protons and is termed the "atomic number" of the element.

J.J. Thomson discovered the electron. Atoms are neutral overall, therefore in Thomson’s ‘plum pudding model’:

  • atoms are spheres of positive charge
  • electrons are dotted around inside
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The weights of a large number of miniature poodles are approximately normally distributed with a mean of 99 kilograms and a stan
Karo-lina-s [1.5K]

Answer:

See explanation below

Explanation:

As I say in the comments, the question is incomplete, however, I will try to answer this by using data that I found on another site.

This is the part of the question that is not here:

If measurements are recorded to the nearest

tenth of a kilogram, find the fraction of these poodles

with weights

(a) over 9.5 kilograms;

(b) of at most 8.6 kilograms;

So, assuming a mean of 8 kg, and 0.9 of standard deviation, let X represents the weight of the poodles

The expression to calculate the fraction of poodle needed is:

Z = X - u / d

u: weight of the large number of poodle

d: standard deviation

Replacing data of a) wer have:

Z = 9.5 - 8 / 0.9

Z = 1.67

With this value, we need to take the value of Z, and see the area under the curve of standard deviation (see table attached)

Therefore:

P (X > 9.5) = P(Z > 1.67) = 0.5 - P (Z < 1.67) = 0.5 - 0.4525 = 0.0475

b) In this part, is the same as part a) so:

Z = 8.6 - 8 / 0.9 = 0.67

The value for area in the curve is 0.2486 so:

P = 0.5 + 0.2486 = 0.7486

Hope this helps

8 0
2 years ago
A 9V battery is directly connected to each of 3 LED bulbs. Select the statement that accurately describes this circuit. A) A dir
Likurg_2 [28]
First off, you can cross out alternating current because a 9V battery doesn't give out AC, it gives out solely DC. If the battery is connected to each battery individually, then they are in parallel. So, according to Kirchhoff's Voltage Law, in parallel, V total = V1 = V2= V3..
So I'd say B) !
7 0
2 years ago
Read 2 more answers
If Pete ( mass=90.0kg) weights himself and finds that he weighs 30.0 pounds, how far away from the surface of the earth is he
shutvik [7]

Answer: 9938.8 km

Explanation:

1 pound-force = 4.48 N

30.0 pounds-force = 134.4 N

The force of gravitation between Earth and object on the surface of is given by:

F = \frac{GMm}{R^2} = mg

Where M is the mass of the Earth, m is the mass of the object, R (6371 km) is the radius of the Earth.

At height, h above the surface of the Earth, the weight of the object:

(mg)'= \frac{GMm}{(R+h)^2}

we need to find "h"

taking the ratio of two:

\frac{mg}{(mg)'}=\frac{(R+h)^2}{R^2}\\ \Rightarrow \frac{90kg \times 9.8 m/s^2}{134.4 N}=\frac{(R+h)^2}{R^2}\\ \Rightarrow 6.56 R^2= (R+h)^2 \Rightarrow h= (2.56-1)R\\ \Rightarrow h = 1.56 R = 1.56 \times 6371 km = 9938. 8 km

Hence, Pete would weigh 30 pounds at 9938.8 km above the surface of the Earth.

5 0
2 years ago
A cylindrical wire has a resistance R and resistivity ρ. If its length and diameter are BOTH cut in half, what will be its resis
snow_lady [41]

Answer:

The resistance will be 2×R

Explanation:

We note that the resistivity of a cylindrical wire is given by the following relation;

\rho = \frac{RA}{L}

Where:

ρ = Resistivity of the wire

R = The wire resistance

A = Cross sectional area of the wire = π·D²/4

L = Length  of the wire

Rearranging, we have;

R= \frac{\rho L}{A}

If the length and the diameter are both cut in half, we have;

L₂ = L/2

A₂ =π·D₂²/4 = \pi \cdot \left (\frac{D}{2}   \right )^{2} \times \frac{1}{4}  = \pi \cdot \frac{D^{2}}{16} = A/4

Therefore, the new resistance, R₂ can be expressed as follows;

R_2= \frac{\rho \frac{L}{2} }{\frac{A}{4} } = \rho \frac{L}{2} \times \frac{4}{A} = 2 \times  \frac{\rho L}{A}

Hence, the new resistance R₂ =  2×R, that is the resistance will be doubled.

8 0
2 years ago
Find the current that flows in a silicon bar of 10-μm length having a 5-μm × 4-μm cross-section and having free-electron and hol
klasskru [66]

The current flowing in silicon bar is 2.02 \times 10^-12 A.

<u>Explanation:</u>

Length of silicon bar, l = 10 μm = 0.001 cm

Free electron density, Ne = 104 cm^3

Hole density, Nh = 1016 cm^3

μn = 1200 cm^2 / V s

μр = 500 cm^2 / V s

The total current flowing in the bar is the sum of the drift current due to the hole and the electrons.

J = Je + Jh

J = n qE μn + p qE μp

where, n and p are electron and hole densities.

J = Eq (n μn + p μp)

we know that E = V / l

So, J = (V / l) q (n μn + p μp)

     J = (1.6 \times 10^-19) / 0.001 (104 \times 1200 + 1016 \times 500)

     J = 1012480 \times 10^-16 A / m^2.

or

J = 1.01 \times 10^-9 A / m^2

Current, I = JA

A is the area of bar, A = 20 μm = 0.002 cm

I = 1.01 \times 10^-9 \times 0.002 = 2.02 \times 10^-12

So, the current flowing in silicon bar is 2.02 \times 10^-12 A.  

6 0
2 years ago
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