Answer:
Psm = 30.66 [Psig]
Explanation:
To solve this problem we will use the ideal gas equation, recall that the ideal gas state equation is always worked with absolute values.
P * v = R * T
where:
P = pressure [Pa]
v = specific volume [m^3/kg]
R = gas constant for air = 0.287 [kJ/kg*K]
T = temperature [K]
<u>For the initial state</u>
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P1 = 24 [Psi] + 14.7 = 165.47[kPa] + 101.325 = 266.8 [kPa] (absolute pressure)
T1 = -2.6 [°C] = - 2.6 + 273 = 270.4 [K] (absolute Temperature)
Therefore we can calculate the specific volume:
v1 = R*T1 / P1
v1 = (0.287 * 270.4) / 266.8
v1 = 0.29 [m^3/kg]
As there are no leaks, the mass and volume are conserved, so the volume in the initial state is equal to the volume in the final state.
V2 = 0.29 [m^3/kg], with this volume and the new temperature, we can calculate the new pressure.
T2 = 43 + 273 = 316 [K]
P2 = R*T2 / V2
P2 = (0.287 * 316) / 0.29
P2 = 312.73 [kPa]
Now calculating the manometric pressure
Psm = 312.73 -101.325 = 211.4 [kPa]
And converting this value to Psig
Psm = 30.66 [Psig]
The mass of Mg-24 is 24.30506 amu, it contains 12 protons and 12 neutrons.
Theoretical mass of Mg-24:
The theoretical mass of Mg-24 is:
Hydrogen atom mass = 12 × 1.00728 amu = 12.0874 amu
Neutron mass = 12 x 1.008665 amu = 12.104 amu
Theoretical mass = Hydrogen atom mass + Neutron mass = 24.1913 amu
Note that the mass defect is:
Mass defect = Actual mass - Theoretical mass : 24.30506 amu- 24.1913 amu= 0.11376 amu
Calculating the binding energy per nucleon:
So approximately 4.41294 Mev/necleon
Answer:
1176.01 °C
Explanation:
Using Ohm's law,
V = IR................. Equation 1
Where V = Voltage, I = current, R = Resistance when the bulb is on
make R the subject of the equation
R = V/I.................. Equation 2
R = 4.3/0.32
R = 13.4375 Ω
Using
R = R'(1+αΔθ)............................. Equation 3
Where R' = Resistance of the bulb at 20°, α = Temperature coefficient of resistivity, Δθ = change in temperature
make Δθ the subject of the equation
Δθ = (R-R')/αR'.................. Equation 4
Given: R = 13.4375 Ω, R' = 1.6 Ω, α = 6.4×10⁻³ K⁻¹
Substitute into equation 4
Δθ = (13.4375-1.6)/(1.6×0.0064)
Δθ = 11.8375/0.01024
Δθ = 1156.01 °C
But,
Δθ = T₂-T₁
T₂ = T₁+Δθ
Where T₂ and T₁ = Final and initial temperature respectively.
T₂ = 20+1156.01
T₂ = 1176.01 °C
Answer:
The moment of inertia is I = 0.126*R^2*M
Explanation:
We can calculate the moment of inertia of an object that starts from rest and has a final velocity using the energy conservation equation, as follows:
Ek1 + Ep1 = Ek2 + Ep2, where
Ek1 = kinetic energy of the object before to roll down
Ep1 = potential energy of the object
Ek2 = kinetic energy when the object comes down
Ep2 = potential energy of the object at the bottom
We have the follow:
Ek1 = 0
Ep1 = M*g*h
Ek2 = ((I*w)/2) + ((M*v^2)/2)
Ep2 = 0
Replacing values:
0 + M*g*h = ((I*w)/2) + ((M*v^2)/2) + 0
where:
M = mass of the object
g = gravitational acceleration
I = moment of the inertia
w = angular velocity = v/R
h = height
M*g*h = ((1/2) * I * (v^2/R^2)) + ((M*v^2)/2)
M*9.8*2 = (I*(5.9^2)/(2*R^2)) + ((5.9^2 * M)/2)
19.6 * M = ((17.4*I)/R^2) + 17.4*M
Clearing I, we have:
I = 0.126*R^2*M
