Answer: 0.43 V
Explanation:
L = [μ(0) * N² * A] / l
Where
L = Inductance of the solenoid
N = the number of turns in the solenoid
A = cross sectional area of the solenoid
l = length of the solenoid
7.3*10^-3 = [4π*10^-7 * 450² * A] / 0.24
1.752*10^-3 = 4π*10^-7 * 202500 * A
1.752*10^-3 = 0.255 * A
A = 1.752*10^-3 / 0.255
A = 0.00687 m²
A = 6.87*10^-3 m²
emf = -N(ΔΦ/Δt).........1
L = N(ΔΦ/ΔI) so that,
N*ΔΦ = ΔI*L
Substituting this in eqn 1, we have
emf = - ΔI*L / Δt
emf = - [(0 - 3.2) * 7.3*10^-3] / 55*10^-3
emf = 0.0234 / 0.055
emf = 0.43 V
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:
I dont see a picture where is it?
Explanation:
I cannot see anything L question Luestion
I assume that the force of 20 N is applied along the direction of motion and was applied for the whole 6 meters, the formula of work is this; Work = force * distance * cosθ where θ is zero degrees. Plugging in the data to the formula; Work = 20 N * 6 m * cos 0º.
Work = 20 N * 6 m * 1
Work = 120 Nm
Work = 120 joules
Hope this helps!
