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Mila [183]
3 years ago
13

A beam of protons is directed in a straight line along the +z direction through a region of space in which there are crossed ele

ctric and magnetic fields. If the electric field is 800 V/m in the −y direction and the protons move at a constant speed of 105 m/s, what must be the magnitude and direction of the magnetic field such that the beam of protons continues along its straight-line trajectory? magnitude .008 Correct: Your answer is correct. T direction
Physics
1 answer:
IceJOKER [234]3 years ago
7 0

Answer:

B = 7.6 T   direction of + x

Explanation:

For the proton beam to continue in the same direction the electric and magnetic forces must be equal

              F_{m} - F_{e} = 0

              F_{m} = F_{e}

              Fm = q E

             

The electric force is in the direction of the electric field because it is the charge of the positive proton, the electric force goes in the direction of –y, therefore, the magnetic force cancels this force must go in the direction of + y

 The magnetic force is

             F_{m} = q v x B = q v B sin θ

             θ = 90

             B = q E / q v

             B = E / v

             B = 800/105

             B = 7.6 T

To find the direction of the magnetic field we use the right hand rule, the thumb goes in the direction of the proton velocity, the fingers extended in the direction of the magnetic field and the palm is the direction of force, for a positive charge.

Thumb goes in the direction of the + z axis

Palm in the direction of +y

Fingers point in the direction of + x

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Vector A⃗ points in the negative y direction and has a magnitude of 5 km. Vector B⃗ has a magnitude of 15 km and points in the p
Alexxandr [17]

Answer:

magnitude of A − B =  15.81 km

Explanation:

Vector A points in the negative y-direction and has a magnitude of 5 km. Vector B points in the positive x-direction and has a  magnitude of 15 km.

According to Cartesian coordinate system, the resultant will start either from tail of A and ends at head of B and vice-versa.

A(0,-5)

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A - B = (-15 i - 5 j )

Magnitude of the vector is given by

|A - B| = \sqrt{(-15)^{2}+(-5)^{2}}

|A - B| = \sqrt{250}

|A - B| = 15.81 km

7 0
3 years ago
A coin is dropped into a wishing well. It takes 1.1 seconds for a splash to be heard. Calculate the depth of the wishing well
Zigmanuir [339]

Answer:

If the wishing well was in a vacuum, then s=ut + 0.5 a t^2 (s=distance, ... wishing well if you drop a coin into it and hear the splash 10 seconds

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8 0
3 years ago
Use the data provided to calculate the gravitational potential energy of each cylinder mass. Round your answers to the nearest t
Goryan [66]

Answer: 14.7kJ, 29.4kJ, 44.1kJ

Explanation:

<em>The gravitational potential energy is the energy that a body or object possesses, due to its position in a gravitational field.  </em>

<em />

In the case of the Earth, in which the gravitational field is considered constant, the value of the gravitational potential energy U_{p} will be:  

U_{p}=mgh  

Where m is the mass of the object, g=9.8m/s^{2} the acceleration due gravity and h=500m the height of the object.  

Knowing this, let's begin with the calculaations:

For m=3kg

U_{p}=(3kg)(9.8m/s^{2})(500m)  

U_{p}=14700J=14.7kJ  

For m=6kg

U_{p}=(6kg)(9.8m/s^{2})(500m)  

U_{p}=29400J=29.4kJ  

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U_{p}=(9kg)(9.8m/s^{2})(500m)  

U_{p}=44100J=44.1kJ  

6 0
3 years ago
Mary and her younger brother Alex decide to ride the carousel at the State Fair. Mary sits on one of the horses in the outer sec
GaryK [48]

Mary and her younger brother Alex decide to ride the carousel at the State Fair, Mary's and Alex's  angular speed M and tangential speed vM is mathematically given as

Mary's and Alex's  angular speed=1.43

Tangential speed mary=3.22 m/s

Tangential speed alex =2.260m/s

<h3>What is Mary's and Alex's angular speed M and tangential speed vM?</h3>

Generally, the equation for angular speed is mathematically given as

w=2\pi /T\\\\Therefore\\\\w=2\pi/3.9

w = 1.61 rev/see 3.9

Centripetal acc mary = v^2/r

Centripetal acc mary  = w^2r

Centripetal acc mary = w^2x 2m

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Therefore

\frac{(ac) mary }{(ac) plex}= 1.43

Hence

tang. speed V=Wr

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Read more about Speed

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8 0
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

7 0
2 years ago
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