Answer:
110 m
Explanation:
The geometry of the problem can be modeled by a right triangle. The height of the cliff forms one leg, and the distance to a buoy forms the other leg. The angle of depression is opposite the height of the cliff. An appropriate trig relation is ...
Tan = Opposite/Adjacent
Solving for the Adjacent side (the distance to the buoy), we find ...
Adjacent = Opposite/Tan
distance to buoy 1 = (80 m)/tan(23°) ≈ 188.468 m
distance to buoy 2 = (80 m)/tan(15°) ≈ 298.564 m
Then the distance between the buoys is ...
298.564 -188.468 m ≈ 110 m
The buoys are about 110 meters apart.
Answer:
a) 0.39795 kJ/K
b) 79.589.37 kJ
Explanation:
m = Mass of air = 2 kg
Temperature = 200 K
P₁ = Initial pressure = 300 kPa
P₂ = Final pressure = 600 kPa
R = mass-specific gas constant for air = 287.058 J/kgK
a) For isentropic process
∴ Entropy is generated in the process is 0.39795 kJ/K
b)
∴ Amount of lost work is 79.589.37 kJ
Answer:
I hope this helps you
Explanation:
किसी की जान बचाने के लिए झूठ बोलना सत्य से अधिक उपयोगी है । अन्यायपूर्ण आज्ञा का विरोध आज्ञापालन से श्रेष्ठ है ।