Incomplete question as the charge density is missing so I assume charge density of 3.90×10^−12 C/m².The complete one is here.
An electron is released from rest at a distance of 0 m from a large insulating sheet of charge that has uniform surface charge density 3.90×10^−12 C/m² . How much work is done on the electron by the electric field of the sheet as the electron moves from its initial position to a point 3.00×10−2 m from the sheet?
Answer:
Work=1.06×10⁻²¹J
Explanation:
Given Data
Permittivity of free space ε₀=8.85×10⁻¹²c²/N.m²
Charge density σ=3.90×10⁻¹² C/m²
The electron moves a distance d=3.00×10⁻²m
Electron charge e=-1.6×10⁻¹⁹C
To find
Work done
Solution
The electric field due is sheet is given as
E=σ/2ε₀

Now we need to find force on electron

Now for Work done on the electron
Would you be able to post the gel electrophoresis results?
Yes, that is truly amazing.
They significantly reduced the noise's intensity by 90%, that might be proper reaction.
<h3>What is noise?</h3>
- Unwanted sound that is regarded loud, unpleasant, or disruptive to hearing is called noise.
- Physically speaking, there is no difference between unwanted sound and desired sound because both are vibrations traveling through a medium like air or water.
- When the brain receives and interprets a sound, a difference occurs.
- Decibels are used to measure sound (dB).
- A motorcycle engine operating is roughly 95 dB louder than regular conversation, which is around 60 dB louder than a whisper.
- Your hearing may begin to be harmed if exposed to noise over 70 dB for an extended period of time.
- Your ears can suffer instant damage from loud noise above 120 dB.
<h3>What is sound intensity?</h3>
- The power carried by sound waves per unit area in a direction perpendicular to that region is known as sound intensity or acoustic intensity.
- The watt per square meter is the SI measure of intensity, which also covers sound intensity.
Learn more about sound here:
brainly.com/question/14595927
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Answer:
Tension T in each rope will be 254.56 N.
Explanation:
From the picture attached,
Weight suspended by the two ropes will be supported by the vertical components of the tension in the ropes.
Vertical components of both the ropes = Tsin(45)° + Tsin(45)°
= 2Tsin(45)°
= 
= 
Since, 
T = 
T = 254.56 N
Therefore, tension T in each rope will be 254.56 N.