Explanation:
it is given that, the linear charge density of a charge, 
Firstly, we can define the electric field for a small element and then integrate for the whole. The very small electric field is given by :
..........(1)
The linear charge density is given by :


Integrating equation (1) from x = x₀ to x = infinity



Hence, this is the required solution.
Do you have the answer choices ?
Answer:
Yes, it is very helpful.
Explanation:
It's helpful since in a cell, plant or animal, there are a lot of different things. It's hard to memorize everything and know what they look like. Using a model can help you memorize everything better and even understand it better. If someone asked me where or what something was in a cell I think I would be able to recognize it better.
I hope this helps!
The kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.
To find the answer, we have to know about the Lorentz transformation.
<h3>What is its kinetic energy as measured in the Earth reference frame?</h3>
It is given that, an alien spaceship traveling at 0.600 c toward the Earth, in the same direction the landing craft travels with a speed of 0.800 c relative to the mother ship. We have to find the kinetic energy as measured in the Earth reference frame, if the landing craft has a mass of 4.00 × 10⁵ kg.

- Let us consider the earth as S frame and space craft as S' frame, then the expression for KE will be,

- So, to
find the KE, we have to find the value of speed of the approaching landing craft with respect to the earth frame. - We have an expression from Lorents transformation for relativistic law of addition of velocities as,

- Substituting values, we get,


Thus, we can conclude that, the kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.
Learn more about frame of reference here:
brainly.com/question/20897534
SPJ4
Answer:
Systematic error can be corrected using calibration of the measurement instrument, while random error can be corrected using an average measurement from a set of measurements.
Explanation:
Random errors lead to fluctuations around the true value as a result of difficulty taking measurements, whereas systematic errors lead to predictable and consistent departures from the true value due to problems with the calibration of your equipment.
Systematic error can be corrected, by calibration of the measurement instrument. Calibration is simply a procedure where the result of measurement recorded by an instrument is compared with the measurement result of a standard value.
Random error can be corrected using an average measurement from a set of measurements or by Increasing sample size.