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alexandr1967 [171]
3 years ago
12

How can the motion of a charged particle be used to distinguish between a magnetic field and an electric field in a certain regi

on? The charged particle follows a spiral path in a magnetic field. The charged particle follows a linear path in a magnetic field. The charged particle follows a spiral path in an electric field
Physics
1 answer:
Helen [10]3 years ago
4 0

Answer:

The charged particle follows a spiral path in a magnetic field.

Explanation:

A charge immersed in a region with an electric field experiences a force that acts along the same direction of the electric field. In particular:

- The force has the same  direction as the electric field if the charge is positive

- The force has the opposite direction as the electric field if the charge is negative

Therefore, a charge moving in an electric field is accelerated along the direction of the electric field.

On the other hand, a charge in motion in a region with a magnetic field experiences a force that acts perpendicular to the direction of the field. This means that a charge in motion in a magnetic field will acquire a circular motion in the plane perpendicular to the direction of the magnetic field.

As a result, if the particle has also a original motion outside this plane, its final motion will consist of:

- A uniform motion along that direction, +

- A circular motion along the plane perpendicular to the field

So, the resultant motion of the particle will be a spiral path. So the correct answer is

The charged particle follows a spiral path in a magnetic field.

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A gasoline engine receives 200 J of energy from combustion and loses 150 J as heat to the exhaust. What is its efficiency? 75% 2
choli [55]
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3 years ago
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A child is holding a wagon from rolling straight back down in a driveway that inclined at 20 degree horizontal. if the wagon wei
Alenkinab [10]

Answer:

F = 51.3°

Explanation:

The component of weight parallel to the inclined plane must be responsible for the rolling back motion of the car. Hence, the force required to be applied by the child must also be equal to that component of weight:

F = Parallel\ Component\ of\ Weight\ of\ Wagon= WSin\theta\\

where,

W = Weight of Wagon = 150 N

θ = Angle of Inclinition = 20°

Therefore,

F = (150\ N)Sin\ 20^o

<u>F = 51.3°</u>

8 0
2 years ago
A horizontal spring with spring constant 130 N/m is compressed 17 cm and used to launch a 2.8 kg box across a frictionless, hori
olasank [31]

Explanation:

The given data is as follows.

        k = 130 N/m,       \Delta x = 17 cm = 0.17 m   (as 1 m = 100 cm)

     mass (m) = 2.8 kg

When the spring is compressed then energy stored in it is as follows.

             Energy = \frac{1}{2}kx^{2}

Now, spring energy gets converted into kinetic energy when the box is launched.

So,    \frac{1}{2}kx^{2} = \frac{1}{2}mv^{2}

   \frac{1}{2} \times 130 \times (0.17)^{2} = \frac{1}{2} \times 2.8 \times v^{2}

          v^{2} = \frac{3.757}{2.8}

                     = 1.34

                v = 1.15 m/sec

Now,

           Frictional force = \mu \times mg

                                    = 0.15 \times 2.8 \times 9.8

                                    = 4.116 N

Also,  Kinetic energy = work done by friction

           \frac{1}{2}mv^{2} = F_{f} \times d

           \frac{1}{2} \times 2.8 \times (1.15)^{2} = 4.116 \times d

             1.8515 = 4.116 \times d

                 d = 0.449 m

Thus, we can conclude that the box slides 0.449 m across the rough surface before stopping.

8 0
3 years ago
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