Given:
u(initial velocity)=0
v(final velocity)= 10 m/s
t= 4 sec
Now we know that
v= u + at
Where v is the final velocity
u is the initial velocity
a is the acceleration measured in m/s^2
t is the time measured in sec
10=0+ax4
a=10/4
a=2.5 m/s^2
As the temperature of water increases, the density of water will decrease.
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]
Answer:
See explanation below
Explanation:
The equation to use for this is the following:
dU = q + w
As the heat is being release, this value is negative, and same here happens with the work done, because it's in the surroundings.
Therefore the change in the energy would be:
dU = -2.59x10^4 - 6.46^4
dU = -9.05x10^4 kJ
