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Phantasy [73]
4 years ago
10

J.J Thompson proposed that atoms were made up of positively charged particles with electrons scattered into them. How did he rea

ch this conclusion?
Physics
1 answer:
Harlamova29_29 [7]4 years ago
8 0
First, J.J Thompson experimented with cathode ray tubes: sealed glass tubes at vaccum (without air inside), with two electrons (cathode and anode). When a high voltage was applied a beam of particles left from the cathode and passed throuhg two charged plates (one negative and one positive).

The beam of particles was deflected toward the positively-charged electric plate, which indeicatedd that the particles were negatively charged.

Given that Thompson experimented with different materials (cathode electrodes), that the results were always the same, and that the mass of the electrons were a small fraction (approximately 1/2000) of the hydrogen atom, he concluded that the negatively particle was part of the atom (a subatomic particle).

Also, given that the atom is neutral, he concluded that there were negative and positive particles in any atom, and he speculated that the negative particles (electrons) were scattered into the positive particles which were way more massive.


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Two particles with charges of 2nC and 5nC are separated by a distance of 3m. The charge 2nC is placed on the left. Find the forc
Vanyuwa [196]

Answer:

1\cdot 10^{-8} N to the left

Explanation:

The magnitude of the electrostatic force between two charges is given by the following equation:

F=k\frac{q_1 q_2}{r^2}

where:

k=9\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the magnitude of the two charges

r is the distance between the two charges

Moreover, the force is:

- Attractive if the charges have opposite sign

- Repulsive if the charges have same sign

In this problem, we have:

q_1=2nC=2\cdot 10^{-9}C is the magnitude of charge 1

q_2=5nC =5\cdot 10^{-9}C is the magnitude of charge 2

r = 3 m is the distance between the two charges

Substituting, we find the force on both charges:

F=\frac{(9\cdot 10^9)(5\cdot 10^{-9})(2\cdot 10^{-9}}{3^2}=1\cdot 10^{-8} N

Here, the two charges are both positive, so the force is repulsive; since the 2 nC charge is on the left, this means that the force on this charge is to the left (away from the 5 nC charge).

5 0
3 years ago
The surface temperature of a planet depends on both the distance of the planet from the Sun and on how the panet's atmosphere di
BaLLatris [955]

Answer:

Aphelion: 6404 W/m2

Perihelion: 14978 W/m2

Explanation:

The solar energy flux depends on the solar power output divided by the surface of a sphere with a radius equal to the distance to the Sun.

\Phi sol = \frac{Psol}{4 * \pi * d^2}

The distances we need are the aphelion and perihelion of Mercury.

Planetary orbits are ellipses. In an ellipse the eccentricity is related to linear eccentricity and the length of the semi major axis:

e = \frac{c}{a}

Where

e: eccentricity

c: linear eccentricity

a: semi major axis

The linear eccentricity is equal to the distance of the focus of the center of the ellipse.

c = a * c =

a = 0.39 AU = 5.83e10 m

c = 5.83e10e * 0.21 = 1.22e10 m

In planetary orbits the Sun is in one of the fucuses. With this we can calculate the prihelion and aphelion as:

Ap = a + c = 5.83e10 + 1.22e10 = 7.05e10 m

Pe = a - c = 5.83e10 - 1.22e10 = 4.61e10 m

And the solar energy fluxes will be:

\Phi Ap = \frac{4e26}{4 * \pi * 7.05e10^2} = 6404 W/m2

\Phi Pe = \frac{4e26}{4 * \pi * 4.61e10^2} = 14978 W/m2

4 0
3 years ago
A force vector has a magnitude of 599 newtons and points at an angle of 40.8° below the positive x axis. What is the x scalar co
ASHA 777 [7]

Answer:

F_{x}=453.44N

Explanation:

Given data

Force F=599 N

Angle α=40.8°

To find

x scalar component

Solution

The Scalar x component can be found by

F_{x}=FCos\alpha  \\F_{x}=599Cos(40.8)\\F_{x}=453.44N

The Scalar y component can be found by

F_{y}=-FSin\alpha  \\F_{y}=-599Sin(40.8)\\F_{y}=-391.4N

3 0
3 years ago
The resistance, R, to electricity of a cylindrical-shaped wire is given by the equation , where p represents the resistivity of
valentina_108 [34]

Answer:

infinity

Explanation:

Given that

Resistance = R

Resistivity = ρ

Length = L

Diameter = d

The resistance of wire R given as

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

A=Area

A=\dfrac{\pi d^2}{4}

Now by putting the value of A

R=\rho\dfrac{L}{\dfrac{\pi d^2}{4}}

R=\rho\dfrac{4L}{\pi d^2}

When d tends to infinity then d² will also tends to infinity.

So when d tends to zero then the resistance tends to infinity.

Therefore answer is ---

infinity

7 0
3 years ago
Read 2 more answers
Two men, Joel and Jerry, each pushes an object that are identical on a horizontal frictionless floor starting from rest. Joel an
Nataliya [291]

Answer:

The work done by Joel is greater than the work done by Jerry.

Explanation:

Let suppose that forces are parallel or antiparallel to the direction of motion. Given that Joel and Jerry exert constant forces on the object, the definition of work can be simplified as:

W = F\cdot \Delta s

Where:

W - Work, measured in joules.

F - Force exerted on the object, measured in newtons.

\Delta s - Travelled distance by the object, measured in meters.

During the first 10 minutes, the net work exerted on the object is zero. That is:

W_{net} = W_{Joel} - W_{Jerry}

W_{net} = F\cdot \Delta s - F\cdot \Delta s

W_{net} = (F-F)\cdot \Delta s

W_{net} = 0\cdot \Delta s

W_{net} = 0\,J

In exchange, the net work in the next 5 minutes is the work done by Joel on the object:

W_{net} = W_{Joel}

W_{net} = F\cdot \Delta s

Hence, the work done by Joel is greater than the work done by Jerry.

7 0
3 years ago
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