Explanation:
Given:
u = 20 m/s
a = 5 m/s^2
v = 30 m/s
t = ?
Use the first kinematic equation of motion:
v = u + at
t = (v - u)/a = 10/5 = 2 seconds
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:
Electric current is defined as the rate of flow of electric charge in a circuit from point one point to another. This is carried by electrically charged particles within the circuit. Current is represented by the symbol I and its unit measured in Amperes. It is therefore related to the voltage and resistance of the circuit. If the current in the circuit reduces, the rate at which the charge and current on the capacitor reduces also proportionally in an exponential manner.
Explanation:
Since a decrease in the flow of current in the circuit is observed, the implication for the rate at which the charge and voltage on the capacitor is also an exponential decrease in the rate of flow with time. This is because the electric current is directly proportional to the electric charge and the time.
That certain change from that ecosystem will require 50 years or longer because big ecosystems need a long time to restablize the living ecosystem.
For part a)
Since the conical surface is not exposed to the radiation coming from the walls only from the circular plate and assuming steady state, the temperature of the conical surface is also equal to the temperature of the circular plate. T2 = 600 K
For part b)
To maintain the temperature of the circular plate, the power required would be calculated using:
Q = Aσ(T₁⁴ - Tw⁴)
Q = π(500x10^-3)²/4 (5.67x10^-8)(600⁴ - 300⁴)
Q = 5410.65 W
