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WARRIOR
1 month ago
8

Write a hypothesis about the effects of magnetic and electric fields. Use the format of "if . . . then . . . because . . ." and

be sure to answer the lesson question: "How can magnetic and electric fields be demonstrated?"
Please help!
Physics
2 answers:
inna [3.1K]1 month ago
7 0
In your hypothesis, the "if" component refers to having an electric or magnetic field present. The subsequent "then" part should signify a detectable attraction or repulsion. The "because" segment should relate to the qualities that these electromagnetic properties exhibit.
kicyunya [3.2K]1 month ago
4 0
The response might contain too long of a message to be effective, but here’s an example: If an electric or magnetic field is present, the materials that generate those fields will influence each other by either attracting or repelling, due to their inherent electric or magnetic characteristics.
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16 days ago
An end of a light wire rod is bent into a hoop of radius r. The straight part of the rod has length l; a ball of mass M is attac
ValentinkaMS [3465]

Answer:

arcsin(\frac{R\mu}{(R+l)\sqrt{\mu^2+1}})

Explanation:

By applying the Law of Sines,

sin \theta = \frac{sin \phi R}{ l + R}

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mg = N\sqrt{\mu^2+1}

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\mu N = mgsin\phi

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15 days ago
A proton is released from rest at the origin in a uniform electric field that is directed in the positive xx direction with magn
Ostrovityanka [3204]
The alteration in potential energy is  \Delta PE = - 3.8*10^{-16} \ J

In the query, it is stated that

  The intensity of the uniform electric field equals E = 950 \ N/C

     The distance the electron covers is  x = 2.50 \ m

Typically, the force exerted on this electron is expressed mathematically as

     F = qE

Where F signifies the force and  q represents the charge of the electron, which is a fixed value of q = 1.60*10^{-19} \ C

    Thus  

      F = 950 * 1.60 **10^{-19}

      F = 1.52 *10^{-16} \ N

Generally, the work-energy theorem is mathematically framed as

          W = \Delta KE

Where W denotes the work done on the electron by the electric field and  \Delta KE  is the change in kinetic energy

Additionally, work done on the electron can also be described as

        W = F* x *cos( \theta )

Where  \theta = 0 ^o assuming that the electron's movement aligns with the x-axis  

        So

             \Delta KE = F * x cos (0)

Inserting values

         \Delta KE = 1.52 *10^{-16} * 2.50 cos (0)

          \Delta KE = 3.8*10^{-16} J

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Where \Delta PE signifies the change  in  potential energy  

Thus  

        \Delta PE = - 3.8*10^{-16} \ J

               

7 0
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Answer:

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