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Sergeeva-Olga [200]

3 years ago

14

A charge enters a uniform magnetic field oriented perpendicular to its path. The field deflects the particle into a circular arc

of radius R. If the charged particle enters the same magnetic field with three times the speed, what will be the radius of the circular arc?

1 answer:

frozen [14]3 years ago

7 0

Answer:

The new radius is three times the initial radius

Explanation:

The magnetic force is given by the equation

F = q v x B

Where the blacks indicate vectors, q is the electric charge, v the velocity of the particle and B the magnetic field. The scalar magnitude of this force is

F = q v B sin θ

Where θ is the angle between the velocity and the magnetic field.

As the force is perpendicular to the velocity, the particle describes a circular motion, using Newton's second law we can find the acceleration that is centripetal.

F = ma

a = v² / r

q v B = mv² / r

R = mv / qB

Let's calculate for the new speed (v₂ = 3v)

R₂ = (m / qB) 3v

R₂ = 3 R

R₂ / R = 3

The new radius is three times the initial radius

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A banked circular highway curve is designed for traffic moving at 58 km/h. The radius of the curve is 201 m. Traffic is moving a

skelet666 [1.2K]

Answer:0.077

Explanation:

Given

banked designed for traffic moving at $58 km/h\approx 16.11 m/s$

Radius of the curve 201 m

Actual traffic velocity = $37 km/h approx 10.27 m/s$

For banking of road

$tan\theta =\frac{v^2}{rg}$

$tan\theta =\frac{16.11^2}{201\times 9.81}$

$\theta =7.49^{\circ}$

Centripetal acceleration is given by

$a=\frac{v^2}{r}$

Taking component of centripetal acceleration

along and perpendicular to surface

$a_{parallel}=\frac{v^2cos\theta }{r}$

$a_{perpendicular}=\frac{v^2sin\theta }{r}$

From FBD

$mgsin\theta -f_s=ma_{parallel}$

$f_s=mgsin\theta -ma_{parallel}$ ----1

where $f_s$ is frictional force

$N-mgcos\theta =ma_{perpedicular}$

$N=mgcos\theta +ma_{perpedicular}$ ----2

and we know coefficient of friction is given by

$\mu =\frac{f_s}{N}$

$\mu =\frac{mgsin\theta -ma_{parallel}}{mgcos\theta +ma_{perpedicular}}$

$\mu =\frac{gsin\theta -\frac{v^2cos\theta }{r}}{gcos\theta +\frac{v^2sin\theta }{r}}$

$\mu =\frac{1.2804-0.5202}{9.726+0.068}$

$\mu =0.077$

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Is chlorine gas a compound

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Awnser question : no

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The correct answer is Meteoroid!

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Part D- Isolating a Variable with a Coefficient In some cases, neither of the two equations in the system will contain a variabl

Shtirlitz [24]

Answer:

D) D = $\frac{5}{4} - \frac{3}{4} \ C$ , E) (C, D) = ( $\frac{17}{7}, \ \frac{-4}{7}$

Explanation:

Part D) two expressions are indicated

3C + 4D = 5

2C +5 D = 2

let's simplify each expression

3C + 4D = 5

4D = 5 - 3C

we divide by 4

D = $\frac{5}{4} - \frac{3}{4} \ C$

The other expression

2C +5 D = 2

2C = 2 - 5D

C = $1 - \frac{5}{2} \ D$

we can see that the correct result is 1

Part E.

It is asked to solve the problem by the substitution method, we already have

D = $\frac{5}{4} - \frac{3}{4} \ C$

we substitute in the other equation

2C +5 D = 2

2C +5 (5/4 - ¾ C) = 2

we solve

C (2 - 15/4) + 25/4 = 2

-7 / 4 C = 2 - 25/4

-7 / 4 C = -17/4

7C = 17

C = $\frac{17}{7}$

now we calculate D

D = $\frac{5}{4} - \frac{3}{4} \ \frac{17}{7}$

D = 5/4 - 51/28

D = $\frac{35-51}{28}$

D = - 16/28

D = $- \frac{4}{7}$

the result is (C, D) = ( $\frac{17}{7}, \ \frac{-4}{7}$ )

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