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katrin2010 [14]
2 years ago
5

The compound magnesium phosphate has the chemical formula Mg3(PO4)2. In this compound, phosphorous and oxygen act together as on

e charged particle, which is connected to magnesium, the other charged particle.
Physics
1 answer:
S_A_V [24]2 years ago
5 0

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.

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022 (part 1 of 4) 10.0 points A ball is thrown vertically upward with a speed of 24.5 m/s. How high does it rise? The accelerati
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1)

Answer:

Part 1)

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Part 2)

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Part 1)

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v_i = 24.5 m/s

so maximum height of the ball is given by

H = \frac{v_i^2}{2g}

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H = 30.6 m

Part 2)

As we know that final speed will be zero at maximum height

so we will have

v_f - v_i = at

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Since the time of ascent of ball is same as time of decent of the ball

so here ball will same time to hit the ground back

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t = 2.5 s

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since the acceleration due to earth will be same during its return path as well as the time of the motion is also same

so here its final speed will be same as that of initial speed

so we have

v_f = 24.5 m/s

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Answer:

a = 9.76 m/s/s

Explanation:

As we know that the object is released from rest

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y = \frac{1}{2}at^2

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Answer:

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d = v_x t

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