Answer:
The answer is A, B, C, D
Explanation:
This is because gravity is the weakest force of the four fundamental forces, so it automatically cancels letter E
#2 - A metal railing rusting in camp weather. Rusting is a chemical change.
We can solve the problem by using the law of conservation of energy:
- at the beginning, all mechanical energy of the object is just kinetic energy: , where m is the mass and v is the velocity
- at the point of maximum height, all mechanical energy of the object is just gravitational potential energy: , where h is the maximum height
Therefore, the conservation of energy becomes:
Re-arranging, we find the maximum height:
Answer:
5.82x10^-3
Explanation:
If the decimal moves right I think it goes negative in the exponent because the starting number is small...or less than 1
I think
Answer:
a) T2 = 133.5°C
x2 = 0.364
b) Qout = 3959.6 kJ
Explanation:
For this exercise we will make two assumptions: the tank is stationary, the kinetic and potential energy are equal to zero. The second assumption is that there are interactions between the two tanks. The contents of both tanks will be a closed system. The energy balance equals:
ΔEsystem = Ein - Eout
-Qout = ΔUA + ΔUB = (m*(u2-u1))A + (m*(u2-u1))B
The steam properties for the two tanks in the initial state can be found in Table A-4 to table A-6 of Cengel:
P1A = 1000 kPa
T1A = 300°C
v1A = 0.25799 m^3/kg
u1A = 2793.7 kJ/kg
T1B = 150°C
x1 = 0.5
vf = 0.001091 m^3/kg
uf = 631.66 kJ/kg
vg = 0.39248 m^3/kg
ufg = 1927.4 kJ/kg
v1B = vf + x1*vfg = 0.001091 + (0.5*(0.39248-0.001091)) = 0.19679 m^3/kg
u1B = uf + x1*ufg = 631.66 + (0.5*1927.4) = 1595.4 kJ/kg
V = VA + VB = mA*v1A + mB*v1B = (2*0.25799) + (3*0.19679) = 1.106 m^3
m = mA + mB = 3+2 = 5 kg
the specific volume will be equal to:
v2 = V/m = 1.106/5 = 0.2213 m^3/kg
With these calculations, we can looking the new properties in the same tables:
P2 = 300 kPa
v2 = 0.2213 m^3/kg
T2 = Tsat, 300 kPa = 133.5°C
x2 =(v2-vf)/(vg-vf) = (0.22127-0.001073)/(0.60582-0.001073) = 0.364
u2 =uf + x2*ufg = 561.11 + (0.364*1982.1) = 1282.8 kJ/kg
-Qout = (2*(1282.8-2793.7)) + (3*(1282.8-1595.4)) = -3959.6 kJ
Qout = 3959.6 kJ