A roller coaster, traveling with an initial speed of 21 m/s, decelerates uniformly at -3.5

show answer Incorrect Answer 33% Part (b) Find the radius of curvature, in meters, of the path of a proton accelerated through t

timofeeve [1]

The question is incomplete. Here is the complete question.

Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric field and then injected into a region of constant magnetic field with a field strength of 0.65T.

part (a): What is the potential difference, in volts, required in the first part of the experiment to accelerate electrons to a speed of 6.2 x 10⁷m/s?

part (b): Find the radius of curvature, in meters, of the path of a proton accelerated trhough this same potential after the proton crosses into the region with the magnetic field.

part (c) what is the ratio of the radii of curvature for a proton and an electron traveling through this apparatus?

Answer: (a) V = - 109.44 x 10² V

(b) $r_{p}=$ 9.95 x 10⁻¹ m

(c) ratio = 1800

Explanation: (a) <u>Potential</u> <u>difference</u> is defined as the energy a charged particle has between two points in a circuit. It is calculated as

$\Delta V=\frac{pe}{q}$

where

pe is potential energy

q is charge

and its unit is joule/coulomb of Volts (V).

To determine potential difference required to accelerate a particle, we have to use the principle that the total energy of a system is conserved and one transforms into the other.

In this case, potential energy is transformed in kinetic energy:

pe = V.q

ke = $\frac{1}{2}m.v^{2}$

so

$V.q=\frac{1}{2} m.v^{2}$

$V=\frac{m.v^{2}}{2q}$

Calculating:

$V=\frac{9.11.10^{-31}(6.2.10^{7})^{2}}{2(-1.6.10^{-19})}$

V = $-109.44$ x 10²V

Potential difference of an electron to have speed of 6.2x10⁷m/s is $-109.44$ x 10²V.

(b) A particle has a circular motion when there is a magnetic force acting on it.

Velocity and magnetic force are always perpendicular to each other. Because of that, there is no work on the particle and so, kinetic energy and speed are constant. Since magnetic force supplies centripetal force:

$F_{mag} = F_{c}$

$qvB=\frac{mv^{2}}{r}$

$r=\frac{mv}{qB}$

The radius of the curvature, for a proton, will be:

$r=\frac{1.67.10^{-27}.6.2.10^{7}}{1.6.10^{-19}.0.65}$

r = 9.95 x 10⁻¹m

The raius of curvature, when it is a proton, is 0.995m.

(c) Radius of curvature, if it was a electron:

$r=\frac{9.11.10^{-31}.6.2.10^{7}}{1.6.10^{-19}.0.65}$

r = 54.33 x 10⁻⁵m

ratio = $\frac{9.95.10^{-1}}{54.33.10^{-5}}$

ratio = 1800

Ratio of radii of curvature is 1800, meaning curvature created when it is a proton is 1800 times bigger than when it is a electron.

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4 years ago

How can the center of gravity of a rigid body be determined?

ipn [44]

To calculate the center of gravity, divide total weight distance moment by total mass of the system. Thus, the center of gravity is 13 meter from left-hand side.

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3 years ago

Gina is driving her car to work, but she’s stopped at a red light. When the light turns green, she presses the gas pedal and acc

Alex777 [14]

The net force of the car is greater than zero in the horizontal direction. If it were not greater thn zero, then the vehicle would remain stationary.

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3 years ago

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In each of the parts of this question, a nucleus undergoes a nuclear decay. Determine the resulting nucleus in each case.

MA_775_DIABLO [31]

A) Francium-223

In an alpha decay, a nucleus decay emitting an alpha particle, which corresponds to a nucleus of helium: so, it consists of 2 protons and 2 neutrons.

$X \rightarrow X' + \alpha$

This means that in the decay:

- The original nucleus loses 2 protons --> so its atomic number Z decreases by 2 units

- The original nucleus loses 2 nucleons (2 protons and 2 neutrons) --> so its mass number A decreases by 4 units

In this example, the original nucleus is Ac (Actinium), with

Z = 89

A = 227

After the decay, it must be

Z - 2 = 89 - 2 = 87

A - 4 = 227 - 4 = 223

We see from the periodict table, Z=87 corresponds to Francium (Fr), so the final nucleus will be francium-223 (the isotope of francium with 223 nucleons).

B) Polonium-211

In a beta-minus decay, a neutron in the nucleus turns into a proton, emitting a fast-moving electron (the beta particle) and an anti-neutrino.

$n \rightarrow p + e^- + \bar{\nu}$

Therefore, in this process:

- The original nucleus gains 1 protons, so its atomic number Z increases by 1 unit

- The original nucleus does not lose/gain nucleons, so its mass number A remains the same

In this example, the original nucleus is Bi (bismuth)-211, with

Z = 83

A = 211

So After the decay, it will be

Z + 1 = 83 + 1 = 84

A = 211

So, the nucleus will be Polonium (Z=84), isotope with 211 nucleons.

C) Neon-22

In a beta-plus decay, a proton in the nucleus turns into a neutron, emitting a fast-moving positron (the beta particle) and a neutrino.

$p \rightarrow n + e^+ +\nu$

Therefore, in this process:

- The original nucleus loses 1 protons, so its atomic number Z decreases by 1 unit

- The original nucleus does not lose/gain nucleons, so its mass number A remains the same

In this example, the original nucleus is Na (sodium)-22, with

Z = 11

A = 22

So After the decay, it will be

Z - 1 = 11 - 1 = 10

A = 22

So, the nucleus will be Neon (Z=10), isotope with 22 nucleons.

D) Technetium-98

In a gamma decay, an unstable nucleus emits a gamma ray:

$X' \rightarrow X + \gamma$

In this process, only energy is released (in the form of gamma ray), so there is no gain/loss of protons/neutrons in the process. This means that:

- The atomic number Z remains constant

- The mass number A remains constant

In this example, we have a nucleus of Tc (Technetium)-98, with

Z = 43

A = 98

These numbers will not change during the decay: this means that after the decay, we will still have a nucleus of Technetium-98.

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3 years ago

A solar collector is a device that absorbs ___________from the sun ?

Gekata [30.6K]

Answer:

SUNLIGHT OR SUNRAYS OR ENERGY OR HEAT

Explanation:

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