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:
The genetic makeup in a organism.
Explanation:
I think the answer is CuF2
When the gold cube is immersed in mercury, the tension in the string in Newtons is 3.142N.
<h3>What is tension?</h3>
Tension is the force acting on the linear object like string, chain or rope due to pulling.
Volume of gold V = mass / density
V = 1.18 /19.3x 10³ =61.1 x 10⁻⁶ m³
Tension in the string after immersing will be
T = [ρ(Gold) -ρ(Hg)] g. V
T =[ 19.3x 10³ - 13.6 x 10³] x 9.81 x 61.1 x 10⁻⁶
T =3.416 N
Thus, the tension in the string is 3.42 N.
Learn more about tension.
brainly.com/question/4087119
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