A closed system of mass 10 kg undergoes a process during which there is energy transfer by work from the system of 0.147 kJ per
1 answer:
Result: -50.005 kJ
Details:
Provided Data
mass of the system = 10 kg
work done = 0.147 kJ/kg
Elevation change
initial speed
Final Speed
Specific internal Energy
according to the first Law of thermodynamics
where KE represents kinetic energy
PE indicates potential energy
U denotes internal Energy
Q = 1.47 + 3.375 - 4.850 - 50
Q = -50.005 kJ
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Let us denote as the highest acceleration of the truck by itself. The force generated by the engine to propel the truck in this scenario is
where m is the mass of the truck alone.
Upon attaching a bus that has double the mass of the truck, the total mass of the entire system (truck+bus) becomes (m+2m)=3m. In this situation, the force generated by the engine is
where a2 represents the new acceleration.
Since the engine remains unchanged, the force generated stays the same, allowing us to set the force equations equal to each other:
and solving for a2 gives us:
Answer:
The required energy remains identical in both scenarios since the specific heat capacity (Cp) does not change with varying pressure.
Explanation:
Given;
initial temperature, t₁ = 50 °C
final temperature, t₂ = 80 °C
Temperature change, ΔT = 80 °C - 50 °C = 30 °C
Pressure for scenario one = 1 atm
Pressure for scenario two = 3 atm
The energy needed in both scenarios is expressed as;
Where;
Cp denotes specific heat capacity, which only varies with temperature and remains unaffected by pressure.
Hence, the energy required remains the same for both scenarios since specific heat capacity (Cp) is pressure-independent.
Answer:
Explanation:
a )
Each blade resembles a rod with its axis positioned near one end.
The moment of inertia for one blade is:
= 1/3 x m l²
where m stands for the mass of the blade
l represents the length of each blade.
Total moment of inertia for 3 blades is:
= 3 x x m l²
ml²
2 )
Details provided include:
m = 5500 kg
l = 45 m
Substituting these values produces:
moment of inertia of one blade:
= 1/3 x 5500 x 45 x 45
= 37.125 x 10⁵ kg.m²
Moment of inertia for 3 blades:
= 3 x 37.125 x 10⁵ kg.m²
= 111.375 x 10⁵ kg.m²
c )
Angular momentum
= I x ω
I denotes the moment of inertia of the turbine
ω symbolizes angular velocity
ω = 2π f
f indicates the rotational frequency of the blades
d )
We have I = 111.375 x 10⁵ kg.m² (Calculated)
f = 11 rpm (revolutions per minute)
= 11 / 60 revolutions per second
ω = 2π f
= 2π x 11 / 60 rad / s
Calculating angular momentum yields
= I x ω
111.375 x 10⁵ kg.m² x 2π x 11 / 60 rad / s
= 128.23 x 10⁵ kgm² s⁻¹.