The water molecules with a slower speed are escaping
Answer:
(a) ΔP=0.0245 kPa
(b) P=9.14 kPa
(c)ΔP=0.0245 kPa
Explanation:
(a) As it is perfect gas we can use
(P₁V₁)/T₁=(P₂V₂)/T₂
Since this constant volume so
P₁/T₁=P₂/T₂
T₂ is change in temperature
T₂=1.00+273.16
T₂=274.16 K
ΔP=6.71449-6.69
ΔP=0.0245 kPa
(b) As
(c) Same steps as in part (a)
ΔP=9.164-9.14
ΔP=0.0245kPa
Answer:
Explanation:
Given a square side loop of length 10cm
L=10cm=0.1m
Then, Area=L²
Area=0.1²
Area=0.01m²
Given that, frequency=60Hz
And magnetic field B=0.8T
a. Flux Φ
Flux is given as
Φ=BA Sin(wt)
w=2πf
Φ=BA Sin(2πft)
Φ=0.8×0.01 Sin(2×π×60t)
Φ=0.008Sin(120πt) Weber
b. EMF in loop
Emf is given as
EMF= -N dΦ/dt
Where N is number of turns
Φ=0.008Sin(120πt)
dΦ/dt= 0.008×120Cos(120πt)
dΦ/dt= 0.96Cos(120πt)
Emf=-NdΦ/dt
Emf=-0.96NCos(120πt). Volts
c. Current induced for a resistance of 1ohms
From ohms law, V=iR
Therefore, Emf=iR
i=EMF/R
i=-0.96NCos(120πt) / 1
i=-0.96NCos(120πt) Ampere
d. Power delivered to the loop
Power is given as
P=IV
P=-0.96NCos(120πt)•-0.96NCos(120πt)
P=0.92N²Cos²(120πt) Watt
e. Torque
Torque is given as
τ=iL²B
τ=-0.96NCos(120πt)•0.1²×0.8
τ=-0.00768NCos(120πt) Nm
