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barxatty [35]
2 years ago
8

An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m. Part A How m

uch work is required to bring an additional electron from infinity to the origin? Express your answer with the appropriate units. Part B If, instead of the second electron coming in from infinity, it is initially at x = + 20.00 m on the axis and is given an initial velocity of 100.0 m/s toward the origin, does it reach the origin? Part C How close to the origin does it come?
Physics
1 answer:
r-ruslan [8.4K]2 years ago
4 0

PART A)

Electrostatic potential at the position of origin is given by

V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}

here we have

q_1 = 1.6 \times 10^{-19} C

q_2 = -1.6 \times 10^{-19} C

r_1 = r_2 = 1 m

now we have

V = \frac{Ke}{r} - \frac{Ke}{r}

V = 0

Now work done to move another charge from infinite to origin is given by

W = q(V_f - V_i)

here we will have

W = e(0 - 0) = 0

so there is no work required to move an electron from infinite to origin

PART B)

Initial potential energy of electron

U = \frac{Kq_1e}{r_1} + \frac{kq_2e}{r_2}

U = \frac{9\times 10^9(-1.6\times 10^{-19}(-1.6 \times 10^{-19})}{19} + \frac{9\times 10^9(1.6\times 10^{-19}(-1.6 \times 10^{-19})}{21}

U = (2.3\times 10^{-28})(\frac{1}{19} - \frac{1}{21})

U = 1.15\times 10^{-30}

Now we know

KE = \frac{1}{2}mv^2

KE = \frac{1}{2}(9.1\times 10^{-31}(100)^2

KE = 4.55 \times 10^{-27} kg

now by energy conservation we will have

So here initial total energy is sufficient high to reach the origin

PART C)

It will reach the origin

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Answer:

<em>Total momentum is conserved</em>

Explanation:

<u>Conservation of Momentum </u>

The momentum is a physical magnitude that measures the product of the object's velocity by its mass. The total momentum of a system is the sum of all its components' individual momentums. The two-bear system starts with a total moment of  

p=m_1v_1+m_2v_2=(10)(10)+(10)(-50)=50\ kg.m/s

When both bears stick together, the total mass is 20 kg, and the new momentum is

p'=(20)(2.5)=50 \ kg.m/s

We have assumed both bears move to the right after the collision. In this situation, the total momentum is conserved

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3 years ago
How would a chart of mid-17th century religious beliefs differ from this chart?
tino4ka555 [31]

Answer:

The correct answer is B)  Fewer people would identify as an atheist because people were not willing to share alternative religious beliefs publically.

Explanation:

A chart of mid-17th-century religious beliefs differs from this chart in that Fewer people would identify as an atheist because people were not willing to share alternative religious beliefs publically.

In the 1600s, people did not have total freedom of speech if they had any. The church had a tremendous influence in the life of people and religious beliefs defined societies and families. The Church exerted its power and influence in many aspects of the people's lives and something out of the purview of the church or different to the religious beliefs of the Church was considered to be sacrilegious. The Church prosecuted people for being against the Church, so people of that time preferred to say that they were religious people supporting the church. Being an atheist was not really an option in the 17th century.

<u><em>Hope this helps!!</em></u>

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3 years ago
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why did thomson's from experermenting with cathode rays cause a big change in scientific thought about atoms
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Answer:

His results gave the first evidence that atoms were made up of smaller particles.

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3 years ago
The work function (energy needed to remove an electron) of gold is 5.1 eV. Two pieces of gold (at the same potential) are separa
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Answer:

The value of L is 0.3 nm.

Explanation:

Given that,

Energy \phi= 5.1 eV

Transmission probability = 10⁻³

We need to calculate the value of L

We know that,

Formula of tunneling probability

T=Ge^{-2kL}....(I)

Where,

k=\dfrac{\sqrt{2m(E-U)}}{\hbar}....(II)

Where, m = mass of electron

E = \phi+U

\phi = E-U

Put the value in equation (II)

k=\dfrac{\sqrt{2\times9.1\times10^{-31}\times5.1\times1.6\times10^{-19}}}{1.055\times10^{-34}}

k=1.155\times10^{10} m^{-1}

From equation (I)

ln T=-2kLG

L=\dfrac{ln T}{-2kG}

L=\dfrac{ln 10^{-3}}{-2\times1.155\times10^{10}\times1}

L=2.99\times10^{-10}\ m

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Tracy makes a list of energy transformations that occur as an engine runs to power a car.
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3. Chemical energy transforms into gravitational potential energy.

Explanation:

First of all, let's check the definition for each type of energy:

- Chemical energy is the energy stored in the chemical bonds inside compounds and substances

- Thermal energy is the energy due to the random motion of atoms/molecules (heat)

- Electrical energy is the energy related to the motion of electrical charges

- Gravitational potential energy is the energy of an object which is placed at some height above the ground

- Kinetic energy is the energy of an object in motion

Given this definition, we can check each energy transformation to see which one are really involved in the engine used to power the car:

1. Chemical energy transforms into thermal energy. --> YES. This one is present: in fact, the chemical energy contained in the fuel is converted inside the engine into heat (thermal energy)

2. Thermal energy transforms into electrical energy. --> YES. The heat (thermal energy) used in the engine is converted into electrical energy used to turn on the car.

3. Chemical energy transforms into gravitational potential energy. --> NO. There is no gravitational potential energy involved in this example, since the car runs on the ground.

4. Thermal energy transforms into kinetic energy. --> YES. The heat (thermal energy) used in the engine is converted into energy of motion of the car.

Therefore, Tracy should remove option 3) from her list.

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