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.
The speed is 10.5 miles per minute, or 630 miles per hour.
We don't know the velocity, because you didn't tell us anything about the direction I'm going.
It is a concave lens
Have a nice day
Answer:
The minimum speed of the box bottom of the incline so that it will reach the skier is 8.19 m/s.
Explanation:
It is given that,
Mass of the box, m = 2.2 kg
The box is inclined at an angle of 30 degrees
Vertical distance, d = 3.1 m
The coefficient of friction, 
Using the work energy theorem, the loss of kinetic energy is equal to the sum of gain in potential energy and the work done against friction.


W is the work done by the friction.







v = 8.19 m/s
So, the speed of the box is 8.19 m/s. Hence, this is the required solution.
A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.
A car experiences a deceleration (a) of -41.62 m/s² and comes to a stop (final velocity = v = 0 m/s) after 10.99 m (s).
We can calculate the initial velocity of the car (u) using the following kinematic equation.
![v^{2} = u^{2} + 2as\\\\u = \sqrt[]{v^{2}-2as} = \sqrt[]{(0m/s)^{2}-2(-42.61m/s^{2} )(10.99m)} = 30.60m/s](https://tex.z-dn.net/?f=v%5E%7B2%7D%20%3D%20u%5E%7B2%7D%20%2B%202as%5C%5C%5C%5Cu%20%3D%20%5Csqrt%5B%5D%7Bv%5E%7B2%7D-2as%7D%20%3D%20%5Csqrt%5B%5D%7B%280m%2Fs%29%5E%7B2%7D-2%28-42.61m%2Fs%5E%7B2%7D%20%29%2810.99m%29%7D%20%3D%2030.60m%2Fs)
A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.
Learn more: brainly.com/question/14851168