Answer:
The total charge Q of the sphere is .
Explanation:
Given that,
Radius = 5 cm
Charge density
We need to calculate the total charge Q of the sphere
Using formula of charge
Where, = charge density
V = volume
Put the value into the formula
Put the value into the formula
Hence, The total charge Q of the sphere is .
Answer:
the work is done by the gas on the environment -is W= - 3534.94 J (since the initial pressure is lower than the atmospheric pressure , it needs external work to expand)
Explanation:
assuming ideal gas behaviour of the gas , the equation for ideal gas is
P*V=n*R*T
where
P = absolute pressure
V= volume
T= absolute temperature
n= number of moles of gas
R= ideal gas constant = 8.314 J/mol K
P=n*R*T/V
the work that is done by the gas is calculated through
W=∫pdV= ∫ (n*R*T/V) dV
for an isothermal process T=constant and since the piston is closed vessel also n=constant during the process then denoting 1 and 2 for initial and final state respectively:
W=∫pdV= ∫ (n*R*T/V) dV = n*R*T ∫(1/V) dV = n*R*T * ln (V₂/V₁)
since
P₁=n*R*T/V₁
P₂=n*R*T/V₂
dividing both equations
V₂/V₁ = P₁/P₂
W= n*R*T * ln (V₂/V₁) = n*R*T * ln (P₁/P₂ )
replacing values
P₁=n*R*T/V₁ = 2 moles* 8.314 J/mol K* 300K / 0.1 m3= 49884 Pa
since P₂ = 1 atm = 101325 Pa
W= n*R*T * ln (P₁/P₂ ) = 2 mol * 8.314 J/mol K * 300K * (49884 Pa/101325 Pa) = -3534.94 J
Answer:
Explanation:
Given that,.
A house hold power consumption is
475 KWh
Gas used is
135 thermal gas for month
Given that, 1 thermal = 29.3 KWh
Then,
135 thermal = 135 × 29.3 = 3955.5 KWh
So, total power used is
P = 475 + 3955.5
P =4430.5 KWh
Since 1 hr = 3600 seconds
So, the energy consumed for 1hr is
1KW = 1000W
P = energy / time
Energy = Power × time
E = 4430.5 KWhr × 1000W / KW × 3600s / hr
E = 1.595 × 10^10 J
So, using Albert Einstein relativity equation
E = mc²
m = E / c²
c is speed of light = 3 × 10^8 m/s
m = 1.595 × 10^10 / (3 × 10^8)²
m = 1.77 × 10^-7 kg
Then,
1 kg = 10^6 mg
m = 1.77 × 10^-7 kg × 10^6 mg / kg
m = 0.177mg
m ≈ 0.18 mg