Answer:
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Answer:
power developed by the turbine = 6927.415 kW
Explanation:
given data
pressure = 4 MPa
specific enthalpy h1 = 3015.4 kJ/kg
velocity v1 = 10 m/s
pressure = 0.07 MPa
specific enthalpy h2 = 2431.7 kJ/kg
velocity v2 = 90 m/s
mass flow rate = 11.95 kg/s
solution
we apply here thermodynamic equation that
energy equation that is

put here value with
turbine is insulated so q = 0
so here

solve we get
w = 579700 J/kg = 579.7 kJ/kg
and
W = mass flow rate × w
W = 11.95 × 579.7
W = 6927.415 kW
power developed by the turbine = 6927.415 kW
Answer:
116.3 electrons
Explanation:
Data provided in the question:
Time, t = 2.55 ps = 2.55 × 10⁻¹² s
Current, i = 7.3 μA = 7.3 × 10⁻⁶ A
Now,
we know,
Charge, Q = it
thus,
Q = (7.3 × 10⁻⁶) × (2.55 × 10⁻¹²)
or
Q = 18.615 × 10⁻¹⁸ C
Also,
We know
Charge of 1 electron, q = 1.6 × 10⁻¹⁹ C
Therefore,
Number of electrons past a fixed point = Q ÷ q
= [ 18.615 × 10⁻¹⁸ ] ÷ [ 1.6 × 10⁻¹⁹ ]
= 116.3 electrons
OSHA inspections are generally unannounced. In fact, except in four exceptional circumstances when advance notice may be given.
It is a criminal offense for any person to give unauthorized advance notice of an OSHA inspection.