Answer:
Explanation:
To convert gram / centimeter³ to kg / m³
gram / centimeter³
= 10⁻³ kg / centimeter³
= 10⁻³ / (10⁻²)³ kg / m³
= 10⁻³ / 10⁻⁶ kg / m³
= 10⁻³⁺⁶ kg / m³
= 10³ kg / m³
So we shall have to multiply be 10³ with amount in gm / cm³ to convert it into kg/m³
2.33 gram / cm³
= 2.33 x 10³ kg / m³ .
Answer
given,


mass of book = 0.305 Kg
so, from the diagram attached below




computing horizontal component




θ = 62.35°
Answer:
the required mass flow rate is 49484.37 kg/s
Explanation:
Given the data in the question;
we first determine the relation for mass flow rate of water that passes through the turbine;
so the relation for net work on the turbine due to the change in potential energy considering 100% efficiency is;
= m ( Δ P.E )
so we substitute (gh) for ( Δ P.E );
= m (gh)
m =
/ gh
so we substitute our given values into the equation
m = 100 MW / ( 9.81 m/s²) × 206 m
m = ( 100 MW × 10⁶W/MW) / ( 9.81 m/s²) × 206 m
m = 10 × 10⁷ / 2020.86
m = 49484.37 kg/s
Therefore, the required mass flow rate is 49484.37 kg/s