Answer:
the density of the electrum is 14.30 g/cm³
Explanation:
Given that:
The equilibrium fraction of lattice sites that are vacant in electrum = 
Number of vacant atoms = 
the atomic mass of the electrum = 146.08 g/mol
Avogadro's number = 
The Number of vacant atoms = Fraction of lattice sites × Total number of sites(N)
= × Total number of sites(N)
Total number of sites (N) = 
Total number of sites (N) = 
From the expression of the total number of sites; we can determine the density of the electrum;
where ;
= Avogadro's Number
density of the electrum
= Atomic mass
Thus; the density of the electrum is 14.30 g/cm³
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Answer:
c is the answer because we have to double the number
Answer:
The total system active power P = P_1 + P_2 + P_3 = 34.91 KW
Explanation:
Load 1: Active power P_1 = 20 HP = 14.91kW;
Reactive power 
Load 2: Active power P_2 = 20 kW;
Reactive power Q2 = 0 since the load is purely resistive.
Load 3: Active power 
= 0 due to purely capacitiveload
Reactive power 
= -20 Var
a) since all three loads are connected in parallel therefore
The total system active power P = P_1 + P_2 + P_3 = 34.91 KW
Total system reactive power Q = Q_1 + Q_2 + Q_3 = 11.18 + 0 -20 = -8.82 kVar
Since Q = 0, the power factor is unity.
Supply current per phase is given by
Answer what do you want i know a project but what is it on?
Explanation:
