Different speeds of light through two separate media ... and the difference in wavelength caused by the difference ... causes the bending of waves fronts associated with light rays.
AFTER the bend, since the light rays then travel in a different direction, we may also say that the 'velocity' has changed.
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:
Vi = 0.055 m³ = 55 L
Explanation:
From first Law of Thermodynamics, we know that:
ΔQ = ΔU + W
where,
ΔQ = Heat absorbed by the system = 52.5 J
ΔU = Change in Internal Energy = -102.5 J (negative sign shows decrease in internal energy of the system)
W = Work Done in Expansion by the system = ?
Therefore,
52.5 J = - 102.5 J + W
W = 52.5 J + 102.5 J
W = 155 J
Now, the work done in a constant pressure condition is given by:
W = PΔV
W = P(Vf - Vi)
where,
P = Constant Pressure = (0.5 atm)(101325 Pa/1 atm) = 50662.5 Pa
Vf = Final Volume of System = (58 L)(0.001 m³/1 L) = 0.058 m³
Vi = Initial Volume of System = ?
Therefore,
155 J = (50662.5 Pa)(0.058 m³ - Vi)
Vi = 0.058 m³ - 155 J/50662.5 Pa
Vi = 0.058 m³ - 0.003 m³
<u>Vi = 0.055 m³ = 55 L</u>
