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-BARSIC- [3]
2 years ago
7

An electron passes through two rectangular regions that contain uniform magnetic fields, B1 and B2. The field B1 is stronger tha

n the field B2. Each field fills the region completely. How does the speed v1 of the electron in region 1 compare with the speed v2 in region 2?

Physics
1 answer:
gtnhenbr [62]2 years ago
3 0

Answer:

v1 = v2

Explanation:

Given:

- The missing figure is (attached).

- The Magnetic Field B1 > B2

Find:

How does the speed v1 of the electron in region 1 compare with the speed v2 in region 2?

Solution:

- From Lorentz Law we know that the Force that acts on the charge particle is the cross product of Magnetic Field Vector ( B1 or B2 ) and the velocity vector (v1 or v1).

- From the attached figure related to this problem we see that the electron velocity or direction of motion is always parallel to the magnetic field B1&B2.

- The law of cross product for parallel vector is 0. Hence, the Lorentz force acting on the electron is also zero.

- Zero Force means no work is done on the particle by the magnetic field, thus, the change in kinetic energy also zero for conservation of energy to hold.

- The initial and final kinetic energies of the electron is same. Hence, we can conclude that v1 = v2.

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A pendulum of length L=36.1 cm and mass m=168 g is released from rest when the cord makes an angle of 65.4 degrees with the vert
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(a) -0.211 m

At the beginning the mass is displaced such that the length of the pendulum is L = 36.1 cm and the angle with the vertical is

\theta=65.4^{\circ}

The projection of the length of the pendulum along the vertical direction is

L_y = L cos \theta = (36.1 cm)(cos 65.4^{\circ})=15.0 cm

the full length of the pendulum when the mass is at the lowest position is

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So the y-displacement of the mass is

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(b) 0.347 J

The work done by gravity is equal to the decrease in gravitational potential energy of the mass, which is equal to

\Delta U = mg \Delta y

where we have

m = 168 g = 0.168 kg is the mass of the pendulum

g = 9.8 m/s^2 is the acceleration due to gravity

\Delta y = 0.211 m is the vertical displacement of the pendulum

So, the work done by gravity is

W=(0.168 kg)(9.8 m/s^2)(0.211 m)=0.347 J

And the sign is positive, since the force of gravity (downward) is in the same direction as the vertical displacement of the mass.

(c) Zero

The work done by a force is:

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and the displacement

In this situation, the tension in the string always points in a radial direction (towards the pivot of the pendulum), while the displacement of the mass is tangential (it follows a circular trajectory): this means that the tension and the displacement are always perpendicular to each other, so in the formula

\theta=90^{\circ}, cos \theta = 0

and so the work done is zero.

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2 years ago
1/012=1/0.05+1/d' hiiiiiiiiii
klasskru [66]

Correct question is;

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Answer:

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Explanation:

1/0.12 = (1/0.05) + (1/d')

Let's rearrange to get;

(1/d') = (1/0.12) - (1/0.05)

(1/d') = (1/(12/100)) - (1/(5/100))

(1/d') = 100/12 - 100/5

Let's multiply through by 60 to get rid of the denominators on the right side;

> (1/d') = 500 - 1200

> (1/d') = -700

> d' = -1/700

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