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Salsk061 [2.6K]

2 years ago

7

Problem 8: Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric fi

eld and then injected into a region of constant magnetic field with a field strength of 0.65 T.
What is the potential difference in volts required in the first part of the experiment to accelerate electrons to a speed of 6 1 × 107 m/s?

1 answer:

yanalaym [24]2 years ago

8 0

Answer:

10581.59 V

Explanation:

We are given that

Magnetic field=B=0.65 T

Speed of electron= $v=6.1\times 10^7m/s$

Charge on electron, $q=e=1.6\times 10^{-19} C$

Mass of electron, $m_e=9.1\times 10^{-31} kg$

We have to find the potential difference in volts required in the first part of the experiment to accelerate electrons.

$V=\frac{v^2m_e}{2e}$

Where V=Potential difference

$m_e=$ Mass of electron

v=Velocity of electron

Using the formula

$V=\frac{(6.1\times 10^7)^2\times 9.1\times 10^{-31}}{2\times 1.6\times 10^{-19}}$

$V=10581.59 V$

Hence, the potential difference=10581.59 V

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Answer:

7.55 km/s

Explanation:

The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:

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Re-arranging the equation and substituting numbers, we find the orbital velocity:

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3 years ago

You are bungee jumping from a bridge. Initially, while you are falling the slack bungee cord isn’t exerting any forces or torque

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Answer:

he fall movement we see that both the force is different from zero, and the torque is different from zero.

When analyzing the statements the d is true

Explanation:

Let's pose the solution of this problem, to be able to analyze the firm affirmations.

When the person is falling, the weight acts on them all the time, initially the rope has no force, but at the moment it begins to lash it exerts a force towards the top that is proportional to the lengthening of the rope.

The equation for this part is

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As the axis of rotation is located at the top where they jump, there is a torque.

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2 years ago

A man walks along a straight path at a speed of 4 ft/s. A searchlight is located on the ground 6 ft from the path and is kept fo

BARSIC [14]

We are given that,

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We need to find $\frac{d\theta}{dt}$ when $x=8ft$

The equation that relates x and $\theta$ can be written as,

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$\frac{dx}{dt} = \frac{dx}{d\theta} \cdot \frac{d\theta}{dt}$

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Replacing the value of the velocity

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2 years ago

Town A lies 15 km north of town B. Town C lies 10 km west of town A. A small plane flies directly from town B to town C. What is

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Answer:

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Explanation:

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2 years ago

If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle

ad-work [718]

This question is incomplete, the complete question is;

The electric force due to a uniform external electric field causes a torque of magnitude 20.0 × 10⁻⁹ N⋅m on an electric dipole oriented at 30° from the direction of the external field. The dipole moment of the dipole is 7.5 × 10⁻¹² C⋅m.

What is the magnitude of the external electric field?

If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle?

Answer:

- the magnitude of the external electric field is 5333.3 N/C

- the magnitude of the charge on each particle is 3.0 × 10⁻¹² C ≈ 3 nC

Explanation:

Given that;

Torque = 20.0 × 10⁻⁹ N⋅m

dipole moment = 7.5 × 10⁻¹²

∅ = 30°

The moment T of restoring couple is;

T = PEsin∅

E = T/Psin∅

we substitute

E = 20.0 × 10⁻⁹ N⋅m / (7.5 × 10⁻¹²) sin(30°)

E = 20.0 × 10⁻⁹ / 3.75 × 10⁻¹²

E = 5333.3 N/C

Therefore, the magnitude of the external electric field is 5333.3 N/C

The dipole moment is given by the expression;

p = ql

q = p / l

given that l = 2.5 mm = 0.0025 m

we substitute

q = 7.5 × 10⁻¹² / 0.0025

q = 3.0 × 10⁻¹² C ≈ 3 nC

Therefore, the magnitude of the charge on each particle is 3.0 × 10⁻¹² C ≈ 3 nC

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2 years ago