Answer:
from food eaten (i.e contain carnlbohyadrates)
In other words a infinitesimal segment dV caries the charge
<span>dQ = ρ dV </span>
<span>Let dV be a spherical shell between between r and (r + dr): </span>
<span>dV = (4π/3)·( (r + dr)² - r³ ) </span>
<span>= (4π/3)·( r³ + 3·r²·dr + 3·r·(dr)² + /dr)³ - r³ ) </span>
<span>= (4π/3)·( 3·r²·dr + 3·r·(dr)² + /dr)³ ) </span>
<span>drop higher order terms </span>
<span>= 4·π·r²·dr </span>
<span>To get total charge integrate over the whole volume of your object, i.e. </span>
<span>from ri to ra: </span>
<span>Q = ∫ dQ = ∫ ρ dV </span>
<span>= ∫ri→ra { (b/r)·4·π·r² } dr </span>
<span>= ∫ri→ra { 4·π·b·r } dr </span>
<span>= 2·π·b·( ra² - ri² ) </span>
<span>With given parameters: </span>
<span>Q = 2·π · 3µC/m²·( (6cm)² - (4cm)² ) </span>
<span>= 2·π · 3×10⁻⁶C/m²·( (6×10⁻²m)² - (4×10⁻²m)² ) </span>
<span>= 3.77×10⁻⁸C </span>
<span>= 37.7nC</span>
Answer:
They would attract one another
Explanation:
The interaction between two like-charged objects is repulsive. ... Positively charged objects and neutral objects attract each other negatively charged objects and neutral objects attract each other.
The equation that most accurately represents the model of cellular respiration is: C6H12O6 (sugar) + 6O2 (oxygen) = 6CO2 (carbon dioxide) + 6H2O (water) + energy.
<h3>
CELLULAR RESPIRATION:</h3>
Cellular respiration is the process whereby living organisms obtain energy by breaking down food molecules in their cells.
The process of cellular respiration breaks down sugar molecules (glucose) in the presence of oxygen to produce carbon dioxide and water as products, as well as energy in form of ATP.
Therefore, the equation that most accurately represents the model of cellular respiration is: C6H12O6 (sugar) + 6O2 (oxygen) = 6CO2 (carbon dioxide) + 6H2O (water) + energy.
Learn more about cellular respiration at: brainly.com/question/12671790?referrer=searchResults
Option (a) is correct.
Falling objects accelerate as they approach the ground.This is because of the force of gravity acting on the falling objects. so the velocity of these objects increases continuously as they approach the ground. the acceleration acting on the falling objects is a constant ( close to the surface of earth) and is called as acceleration due to gravity denoted by g. value of g=9.8 m/s².
