La vida de Antonio de Nebrija digamen plis

A proton is projected toward a fixed nucleus of charge Ze with velocity vo. Initially the two particles are very far apart. When

11111nata11111 [884]

Answer:

The value is $R_f = \frac{4}{5} R$

Explanation:

From the question we are told that

The initial velocity of the proton is $v_o$

At a distance R from the nucleus the velocity is $v_1 = \frac{1}{2} v_o$

The velocity considered is $v_2 = \frac{1}{4} v_o$

Generally considering from initial position to a position of distance R from the nucleus

Generally from the law of energy conservation we have that

$\Delta K = \Delta P$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K = K__{R}} - K_i$

=> $\Delta K = \frac{1}{2} * m * v_1^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * (\frac{1}{2} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta P = P_f - P_i$

Here $P_i$ is zero because the electric potential energy at the initial stage is zero so

$\Delta P = k * \frac{q_1 * q_2 }{R} - 0$

So

$\frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R} - 0$

=> $\frac{1}{2} * m *v_0^2 [ \frac{1}{4} -1 ] = k * \frac{q_1 * q_2 }{R}$

=> $- \frac{3}{8} * m *v_0^2 = k * \frac{q_1 * q_2 }{R} ---(1 )$

Generally considering from initial position to a position of distance $R_f$ from the nucleus

Here $R_f$ represented the distance of the proton from the nucleus where the velocity is $\frac{1}{4} v_o$

Generally from the law of energy conservation we have that

$\Delta K_f = \Delta P_f$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K_f = K_f - K_i$

=> $\Delta K_f = \frac{1}{2} * m * v_2^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * (\frac{1}{4} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * \frac{1}{16} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance $R_f$ from the nucleus , this is mathematically represented as

$\Delta P_f = P_f - P_i$

Here $P_i$ is zero because the electric potential energy at the initial stage is zero so

$\Delta P_f = k * \frac{q_1 * q_2 }{R_f } - 0$

So

$\frac{1}{2} * m * \frac{1}{8} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R_f }$

=> $\frac{1}{2} * m *v_o^2 [-\frac{15}{16} ] = k * \frac{q_1 * q_2 }{R_f }$

=> $- \frac{15}{32} * m *v_o^2 = k * \frac{q_1 * q_2 }{R_f } ---(2)$

Divide equation 2 by equation 1

$\frac{- \frac{15}{32} * m *v_o^2 }{- \frac{3}{8} * m *v_0^2 } } = \frac{k * \frac{q_1 * q_2 }{R_f } }{k * \frac{q_1 * q_2 }{R } }}$

=> $-\frac{15}{32 } * -\frac{8}{3} = \frac{R}{R_f}$

=> $\frac{5}{4} = \frac{R}{R_f}$

=> $R_f = \frac{4}{5} R$

7 0

3 years ago

3. If there is an atomic mass of 47

DochEvi [55]

Answer:

20

27

Explanation:

Given parameters:

Atomic mass = 47

Atomic number = 20

Unknown:

Number of protons = ?

Number of neutrons = ?

Solution:

To solve this problem,

Atomic mass = number of protons + number of neutrons

Atomic number = number of protons

So;

Number of protons = 20

Now;

Number of neutrons = Atomic mass - number of protons

Number of neutrons = 47 - 20 = 27

4 0

3 years ago

Can you tell from your experiment so far whether the tapes carry a positive charge or a negative charge? Briefly explain your an

Softa [21]

Answer:

Explanation:

Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field. There are two types of electric charge: positive and negative (commonly carried by protons and electrons respectively). Like charges repel each other and unlike charges attract each other. An object with an absence of net charge is referred to as neutral. Early knowledge of how charged substances interact is now called classical electrodynamics, and is still accurate for problems that do not require consideration of quantum effects.

Electric charge is a conserved property; the net charge of an isolated system, the amount of positive charge minus the amount of negative charge, cannot change. Electric charge is carried by subatomic particles. In ordinary matter, negative charge is carried by electrons, and positive charge is carried by the protons in the nuclei of atoms. If there are more electrons than protons in a piece of matter, it will have a negative charge, if there are fewer it will have a positive charge, and if there are equal numbers it will be neutral. Charge is quantized; it comes in integer multiples of individual small units called the elementary charge, e, about 1.602×10−19 coulombs,[1] which is the smallest charge which can exist freely (particles called quarks have smaller charges, multiples of

e, but they are only found in combination, and always combine to form particles with integer charge). The proton has a charge of +e, and the electron has a charge of −e.

An electric charge has an electric field, and if the charge is moving it also generates a magnetic field. The combination of the electric and magnetic field is called the electromagnetic field, and its interaction with charges is the source of the electromagnetic force, which is one of the four fundamental forces in physics. The study of photon-mediated interactions among charged particles is called quantum electrodynamics.

8 0

3 years ago

For a point charge, how does the potential vary with distance from the point charge, r?.

Sladkaya [172]

For a point charge, The potential varies inversely with distance from the point charge r, that is Potential V ∝ $\frac{1}{r}$

A point charge is defined as a charge that neither has mass nor volume, that is it exists as a single point.

The point charge with respect to the electric potential should be expressed in

V = KQ/r

Here we can say that the electric potential should be varied oppositely to the distance

Electric potential is defined as the amount of work needed to bring a charge from infinity to a certain place in opposition to electrostatic force. Only when there is a potential gradient at the conductor's ends do electrons travel across it.

Refer to more about potential here

brainly.com/question/17058027

#SPJ4

3 0

2 years ago

The graph shows the change in position of an object from point A to point B. The x-axis

Solnce55 [7]

hmm i think thats uhh B or A

Explanation:

Thats all i know hope it helps

3 0

3 years ago

La vida de Antonio de Nebrija digamen plis​

La vida de Antonio de Nebrija digamen plis