Answer:
Velocity of the electron at the centre of the ring, 
Explanation:
<u>Given:</u>
- Linear charge density of the ring=

- 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

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

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

Let
be the change in potential Energy given by

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

So the electron will be moving with 
Answer:
Explanation:
<u>1. Name of the variables:</u>

<u>2. Formulae:</u>




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




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
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.
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