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>
The <span>flow of how a cold pack works on a sprained ankle is based on the second law of thermodynamics which states that energy will flow from a higher to a lower temperature. So your body heat will flow to the cold pack in which you will feel the coldness of the pack.</span>
Answer:
Energy consumed by the electric kettle in 9.5 min =Pt=(2.5×10
3
)×(9.5×60)=14.25×10
5
J
Energy usefully consumed =msΔT=3×(4.2×10
3
)×(100−15)=10.71×10
5
where s=4.2J/g
o
C= specific heat of water and boiling point temp=100
o
C
Heat lost =14.25×10
5
−10.71×10
5
=3.54×10
5
Am not really sure but what i see is D
The centripetal acceleration a is 4.32 
10^-4 m/s^2.
<u>Explanation:</u>
The speed is constant and computing the speed from the distance and time for one full lap.
Given, distance = 400 mm = 0.4 m, Time = 100 s.
Computing the v = 0.4 m / 100 s
v = 4 10^-3 m/s.
radius of the circular end r = 37 mm = 0.037 m.
centripetal acceleration a = v^2 / r
= (4 10^-3)^2 / 0.037
a = 4.32 10^-4 m/s^2.
