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natta225 [31]
3 years ago
11

How did Micheal Faraday discover the magnetic theory and when?​

Physics
1 answer:
serious [3.7K]3 years ago
6 0

Answer:

Faraday made his first discovery of electromagnetism in 1821.  He took the work of Oersted and Ampère on the magnetic properties of electrical currents as a starting point and in 1831 achieved an electrical current from a changing magnetic field, a phenomenon known as electromagnetic induction.

Explanation:

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An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea
melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, v=1.37\times10^7\ \rm m/s

Explanation:

<u>Given:</u>

  • Linear charge density of the ring=0.1\ \rm \mu C/m
  • Radius of the ring R=0.2 m
  • Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C

Potential due the ring at a distance x from the centre of the rings is given by

V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V

Let\Delta U be the change in potential Energy given by

\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s

So the electron will be moving with v=1.37\times10^7\ \rm m/s

5 0
3 years ago
A 45.2-kg person is on a barrel ride at an amusement park. She stands on a platform with her back to the barrel wall. The 3.74-m
elena-14-01-66 [18.8K]

Answer:

  • <u><em>1,230N</em></u>

Explanation:

<u>1. Name of the variables:</u>

   f:frequency\\\\ \omega:angular\text{ }speed\\\\ a_c:centripetal\text{ }acceleration\\\\ F_c:centripetal\text{ }force\\ \\ m:mass\\ \\ d:diameter\\ \\ r:radius\\ \\ g:gravitational\text{ }acceleration

<u>2. Formulae:</u>

         f=\dfrac{number\text{ }of\text{ }revolutions}{time}

          \omega=2\pi f

          a_c=\omega^2 r

           F_c=m\times a_c

<u>3. Solution (calculations)</u>

       f=\dfrac{1}{1.65s}=0.\overline{60}s^{-1}

       \omega=2\pi\times0.\overline{60}\approx 3.808rad/s

      a_c=(3.808rad/s)^2\times (3.74/2m)=27.12m/s^2

      F_c=45.2kg\times27.12m/s^2=1,225.67N\approx 1,230N

3 0
3 years ago
How much heat is required to heat 2 kg of water from 25°C to 40°C?
Dominik [7]

Answer:

126000 J

Explanation:

Applying,

Q = cm(t₂-t₁).................. Equation 1

Where Q = Amount of heat, c = specifc heat capacity of water, m = mass of water, t₁ = Initial temperature, t₂ = Final temperature.

From the question,

Given: m = 2 kg, t₁ = 25°C, t₂ = 40°C

Constant: c = 4200 J/kg.°C

Substitute these value into equation 1

Q = 2×4200(40-25)

Q = 2×4200×15

Q = 126000 J

5 0
3 years ago
In a elastic lab what did you change and why?
Vilka [71]

I changed my undershorts.  The elastic on the old ones I put on that day was deteriorated, and it completely failed when I dripped lab coffee on it, causing falldown.

7 0
3 years ago
Read 2 more answers
Part B For this activity, you'll analyze the motion of the marked spot on the skateboarder's leg. For each dropdown labeled “cm”
Anika [276]

Answer:

Explanation:

c. By using the Select Data button and the Select Data Source optionExplanation:A scatter plot is a plot which is used to plot the points of the data on the horizontal and the vertical axis also it depicts how one variable is affected by the another. After preparing the scatter plot to enter the data in the scatter plot we need to use the data button and then data source option so that the data could be entered in the scatter plothence, option c is correct

7 0
2 years ago
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