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neonofarm [45]
3 years ago
14

A charged particle enters into a uniform magnetic field such that its velocity vector is perpendicular to the magnetic field vec

tor. Ignoring the particle's weight, what type of path will the particle follow?
A) follow a spiralpath.
B) move in a straight line.C) move along a parabolic path.
D)follow a circular path.

Physics
1 answer:
Phoenix [80]3 years ago
7 0

Answer:

The particle path will follow

(d) a circular path

Explanation:

When a charged particle having charge of magnitude 'q' enters into a magnetic field such that its velocity vector '\vec{V}' is perpendicular to the direction of the magnetic field '\vec{B}', then it will experience a force, called Lorentz force (\vec{F_{L}}), given by

\vec{F_{L}} = \vec{V} \times \vec{B}

As shown in the figure, the magnetic field is directed perpendicular to the plane and towards the plane (as shown by the circle and 'X'-sign) and the velocity vector is from left to right on the plane.

According to the property of cross-product, the Lorentz force (F_{L}) acting on the particle will be perpendicular to the instantaneous position of the particle, making the path of the particle to be a circular path,as shown in the figure.

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A ball rolls off the end of a horizontal table that is 4 meters off the ground. It is measured that the ball lands 3 meters away
ElenaW [278]

Answer:

The speed at which the ball rolled off the end of the table is 3.3 m/s

Explanation:

Hi there!

Please, see the attached figure for a graphical description of the problem. Notice that the origin of the frame of reference is located at the edge of the table.

The position vector of the ball can be calculated as follows:

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

Where:

r = position vector.

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity.

g = acceleration due to gravity.

When the ball reaches the ground, its position will be:

r final = (3, -4)

Then:

3 = x0 + v0x · t

-4 = y0 + v0y · t + 1/2 · g · t²

Since the origin of the frame of reference is located at the edge of the table, x0 and y0 = 0. v0y is also 0 ( see the initial velocity vector in the figure to elucidate why). Then:

3 m = v0x · t

-4 m = 1/2 · g · t²

We can solve for "t" in the equation of the y-component and use it in the equation of the x-component to obtain v0x:

-4 m = 1/2 · g · t²

-4 m = -1/2 · 9.8 m/s² · t²

8 m / 9.8 m/s² = t²

t = 0.9 s

Then:

3 m = v0x · 0.9s

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v0x = 3.3 m/s

The speed at which the ball roll off the end of the table is 3.3 m/s

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3 years ago
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Answer:

c) 2.02 x 10^16 nuclei

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[A] = Our incognite

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k = ln 2 / Half-life

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k = 1.40x10⁻⁴s⁻¹

t is time = 1.98 x 10^4 s

[A]₀ = 3.21 x 10^17 nuclei

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<h3>c) 2.02 x 10^16 nuclei</h3>
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