If it is a true or false questions, the answer is TRUE.
Ca₁₀(PO₄)₆(OH)₂ or Ca(OH)₂·3Ca₃(PO₄)₂
PO₄³⁻ phosphate ion
OH⁻ oxyhydroxide ion
Ca²⁺ calcium ion
10*(+2) + 6*(-3) + 2*(-1) = 0
10Ca²⁺ 6PO₄³⁻ 2OH⁻
Answer:
3.84 Ω
Explanation:
From the question given above, the following data were obtained:
Electrical power (P) = 150 W
Voltage (V) = 24 V
Resistance (R) =?
P = IV
Recall:
V = IR
Divide both side by R
I = V/R
P = V/R × V
P = V² / R
Where:
P => Electrical power
V => Voltage
I => Current
R => Resistance
With the above formula (i.e P = V²/R), we can calculate resistance as illustrated below:
Electrical power (P) = 150 W
Voltage (V) = 24 V
Resistance (R) =?
P = V²/R
150 = 24² / R
150 = 576 / R
Cross multiply
150 × R = 576
Divide both side by 150
R = 576 / 150
R = 3.84 Ω
Thus, the resistance is 3.84 Ω
Explanation:
Hydrogen (H)
Helium (He)
Lithium (Li)
Beryllium (Be)
Boron (B)
Carbon (C)
Nitrogen (N)
Oxygen (O)
Fluorine (F)
Neon (Ne)
Sodium (Na)
Magnesium (Mg)
Aluminum (Al)
Silicon (Si)
Phosphorus (P)
Sulfur (S)
Chlorine (Cl)
Argon (Ar)
Potassium (K)
Calcium (Ca)
Hope this is correct and helpful
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