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zaharov [31]
3 years ago
9

Two charged particles are projected into a region where a magnetic field is directed perpendicular to their velocities. If the c

harges are deflected in opposite directions, what are the possible relative charges and directions? (Select all that apply.)
Physics
1 answer:
AleksAgata [21]3 years ago
6 0

Answer:

*If the particles are deflected in opposite directions, it implies that their charges must be opposite

*the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.

Explanation:

When a charged particle enters a magnetic field, it is subjected to a force given by

        F = q v x B

where bold letters indicate vectors

   

this expression can be written in the form of a module

        F = qv B sin θ

and the direction of the force is given by the right-hand rule.

In our case the magnetic field is perpendicular to the speed, therefore the angle is 90º and the sin 90 = 1

If the particles are deflected in opposite directions, it implies that their charges must be opposite, one positive and the other negative.

Furthermore, the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.

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Which kinds of objects emit visible light in the electromagnetic spectrum?
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We know the range of wavelength of the visible spectrum is from 400 nm to 780 nm.





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If 3600 j of work is done in 3.0 s what is the power<br>0.00083W<br>1200W<br>3600W<br>11000W
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1200 W

Explanation:

Power is given by the ratio between work done and time taken:

P=\frac{W}{t}

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In this problem, W = 3600 J and t = 3.0 s. Therefore, the power in this exercise is

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A space vehicle of mass m has a speed v. At some instant, it separates into two pieces, each of mass 0.5m. One of the pieces is
Damm [24]

Answer:

W = ½ m v²

Explanation:

In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation

We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved

initial instant. before separation

        p₀ = m v

final attempt. after separation

       p_{f} = m /2  0 + m /2 v_{f}

       p₀ = p_{f}

       m v = m /2 v_{f}

       v_{f}= 2 v

this is the speed of the second part of the ship

now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body

     

initial energy

         K₀ = ½ m v²

final energy

        K_{f} = ½ m/2  0 + ½ m/2 v_{f}²

        K_{f} = ¼ m (2v)²

        K_{f} = m v²

         

the expression for work is

         W = ΔK = K_{f} - K₀

         W = m v² - ½ m v²

         W = ½ m v²

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3 years ago
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