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 answer is 218
Explanation:
Weight = mass * gravitational acceleration
weight is represented by F
F = 25kg (8.7)
(I'm pretty sure that you don't have to include the meters per second/per second thing)
Answer:
Reduce the friction at the surface
Explanation:
If you can reduce the friction between the load and the plane less effort will be required as you are not having to apply effort to overcome friction.
Answer:
Hope it helps for you :)))))
Answer:
A step by step to walk
Explanation:
One- Make sure your shoes are tied so that you dont trip
Two- Make sure your way is a cleared path so you dont fall or even hurt yourself
three- use both set of lets to go in any direction you want.
Four- when walking make sure to try and keep a steedy pace so that both set of legs are going up and down but in harmony
