Answer:
(a) ΔU=747J
(b) γ=1.3
Explanation:
For (a) change in internal energy
According to first law of thermodynamics the change in internal energy is given as
ΔU=Q-W
Substitute the given values
ΔU=970J-223J
ΔU=747J
For(b) γ for the gas.
We can calculate γ by ratio of heat capacities of the gas
γ=Cp/Cv
Where Cp is the molar heat capacity at constant pressure
Cv is the molar heat capacity at constant volume
To calculate γ we first need to find Cp and Cv
So
For Cp
As we know
Q=nCpΔT
Cp=(Q/nΔT)

From relation of Cv and Cp we know that
Cp=Cv+R
Where R is gas constant equals to 8.314J/mol.K
So

So
γ=Cp/Cv
γ=[(37J/mol.K) / (28.687J/mol.K)]
γ=1.3
Answer:
F=X.F=mxq. 3. 1 N/kg=0.5kg g=9.80
Answer:
The voltage across the capacitor is 1.57 V.
Explanation:
Given that,
Number of turns = 10
Diameter = 1.0 cm
Resistance = 0.50 Ω
Capacitor = 1.0μ F
Magnetic field = 1.0 mT
We need to calculate the flux
Using formula of flux

Put the value into the formula


We need to calculate the induced emf
Using formula of induced emf

Put the value into the formula

Put the value of emf from ohm's law





We know that,


We need to calculate the voltage across the capacitor
Using formula of charge


Put the value into the formula


Hence, The voltage across the capacitor is 1.57 V.
Answer:
r=6.05km/hr
z=59.1 degree to the horizontal
Explanation:
A bird is flying east at 5.2 kilometers/hour relative to the air. There's a crosswind blowing at 3.1 kilometers/hour toward the south relative to the ground. What is the bird's velocity relative to the ground? State your answer to one decimal place
can be solved using pythagoras theorem
r^2=o^2+a^2
r^2=5.2^2+3.1^2
r^2=36.65
r=6.1km/hr is te birds velocity relative to the ground
tanz=5.2/3.1
z=tan^-1(5,2/3.1)
z=59.1 degree to the horizontal
The impulse imparted to the shells equals the change in the momentum:
Fav*(Delta t)= Delta m*v.
The mass change is
Delta m= n*m= (89.9shells)*(88.7g)=7.97Kg
So the average force is
F=((v)*(Delta m))/t= ((929)*(7.97))/4.84=1529.78 N
Since the velocity of the shells is much greater than the velocity of the helicopter, there is no need to use relative velocity.