Answer:
a. -8 cm
Explanation:
= distance of the object = 4 cm
= distance of the image = ?
= focal length of the converging lens = 8 cm
using the lens equation
= - 8 cm
The question is incomplete, the complete question is;
The compound magnesium phosphate has the chemical formula Mg3(PO4)2. In this compound, phosphorus and oxygen act together as one charged particle, which is connected to magnesium, the other charged particle. What does the 2 mean in the formula 5Mg3(PO4)2? A. There are two elements in magnesium phosphate. B. There are two molecules of magnesium phosphate. C. There are two magnesium ions in a molecule of magnesium phosphate. D. There are two phosphate ions in a molecule of magnesium phosphate.
Answer:
There are two phosphate ions in a molecule of magnesium phosphate.
Explanation:
The compound magnesium phosphate is an ionic compound. Ionic compounds always consists of two ions, a positive ion and a negative ion.
In this case, the positive ion is Mg^2+ while the negative ion is PO4^3-.
The subscript, 2 after the formula of the phosphate ion means that there are two phosphate ions in each formula unit of magnesium phosphate.
Answer:
I believe that the answer is D. There are drilling platforms all along the coast that are used to drill for natural gas that can be used to generate electricity.
Explanation:
Solar panels use the sun, and that is renewable.
The power plant uses tides and waves, they are renewable.
Windmills use wind, that is renewable.
So, the answer is D.
Answer:
Electrolytes are considered ions when placed in a solution and allow for adequate conduction of particle charges.
Explanation:
Electrolytes are substances that, when are dissolved in solution, separates into electrical positive charges (cations) and electrical negative charges (anions) which are known as ions.
These ions have an adequate capacity to conduct particle charges and, therefore electricity.
Sodium, calcium, phosphate and potassium, are examples of electrolytes.
<u>Hence, the correct answer is:</u>
Electrolytes are considered ions when placed in a solution and allow for adequate conduction of particle charges.
I hope it helps you!
Given:
Area of pool = 3m×4m
Diameter of orifice = 0.076m
Outlet Velocity = 6.3m/s
Accumulation velocity = 1.5cm/min
Required:
Inlet flowrate
Solution:
The problem can be solved by this general formula.
Accumulation = Inlet flowrate - Outlet flowrate
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice
First, we need to convert the units of the accumulation velocity into m/s to be consistent.
Accumulation velocity = 1.5cm/min × (1min/60s)×(1m/100cm)
Accumulation velocity = 0.00025 m/s
We then calculate the area of the pool and the area of the orifice by:
Area of pool = 3 × 4 m²
Area of pool = 12m²
Area of orifice = πd²/4 = π(0.076m)²/4
Area of orifice = 0.00454m²
Since we have all we need, we plug in the values to the general equation earlier
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice
0.00025 m/s × 12m² = Inlet flowrate - 6.3m/s × 0.00454m²
Transposing terms,
Inlet flowrate = 0.316 m³/s
