Answer:
1353.38 Watt
Explanation:
T₁ = Initial temperature of the house = 35°C
T₂ = Final temperature of the house = 20°C
Δt = Time taken to cool the house = 38 min = 38×60 = 2280 s
m = mass of air in the house = 800 kg
Cv = Specific heat at constant volume = 0.72 kJ/kgK
Cp = Specific heat at constant pressure = 1.0 kJ/kgK
Heat removed
q = mCvΔT
⇒q = 800×720×(35-20)
⇒q = 8640000 J
Average rate of hear removal
∴ Power drawn by the air conditioner is 1353.38 Watt
Answer:
20 5/6 sec
Explanation:
To find the solution we divide 5000 by 240
However, when you see a problem, always try to simplify
5000/240=500/24=250/12=125/6
Now the division is much easier
20 5/6 sec
Answer:
<h2>
a) Q = 0.759µC</h2><h2>
b) E = 39.5µJ</h2>
Explanation:
a) The charge Q on the positive charge capacitor can be gotten using the formula Q = CV
C = capacitance of the capacitor (in Farads )
V = voltage (in volts) = 100V
C = ∈A/d
∈ = permittivity of free space = 8.85 × 10^-12 F/m
A = cross sectional area = 600 cm²
d= distance between the plates = 0.7cm
C = 8.85 × 10^-12 * 600/0.7
C = 7.59*10^-9Farads
Q = 7.59*10^-9 * 100
Q = 7.59*10^-7Coulombs
Q = 0.759*10^-6C
Q = 0.759µC
b) Energy stored in a capacitor is expressed as E = 1/2CV²
E = 1/2 * 7.59*10^-9 * 100²
E = 0.0000395Joules
E = 39.5*10^-6Joules
E = 39.5µJ
