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Rus_ich [418]
3 years ago
10

A satellite dish is being designed so that it can pick up radio waves coming from space. The satellite dish will be in the shape

of a parabola and will be positioned above the ground such that its focus is 40 ft above the ground. Using the ground as the x-axis, where should the base of the satellite be positioned? Which equation best describes the equation of the satellite?
Chemistry
1 answer:
pychu [463]3 years ago
4 0

Answer:

the base of the satellite should be positioned at (0,20)

the best model which describes the equation of the satellite is \mathbf{y = \dfrac{x^2}{80} + 20}

Explanation:

From the information given:

A satellite dish is designed in such a way that it can pick up radio waves coming from the space.

This satellite dish is designed in the shape of a parabola.

Location of the satellite is situated above the ground

Focus of the satellite is 40 ft above the ground

Using the ground as the x - axis

The objective here is to determine where should the base of the satellite be positioned?     &

Which equation best describes the equation of the satellite?

Using the ground as the x - axis;

Let say the x- axis start from the origin (0,0)

Then;

The Focus of the satellite which is 40 ft above the ground will be (0,40)

The position of the base of the satellite will be the vertex (h,k) which is at an equidistant position between the ground and the focus.

(h,k) = (0,20)

The model which best describes the equation of the satellite is as follows :

(x - h)^2 = 4a(y - k) \\ \\ where ; \\ \\ (h,k) = (0,20) , \\ a = 20(x - 0)^2 = 4(20)(y - 20)\\ \\x^2 = 80( y - 20)\\ \\x^2= 80y - 1600\\ \\y = \dfrac{x^2}{80} + 20

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Solutions of sulfuric acid and lead(II) acetate react to form solid lead(II) sulfate and a solution of acetic acid. 4.90 g of su
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Answer:

Mass H2SO4 = 3.42 grams

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Explanation:

Step 1: Data given

Mass of sulfuric acid = 4.90 grams

Molar mass of sulfuric acid = 98.08 g/mol

Mass of lead acetate = 4.90 grams

Molar mass of lead acetate = 325.29 g/mol

Step 2: The balanced equation

H2SO4 + Pb(C2H3O2)2 → PbSO4 + 2CH3COOH

Step 3: Calculate moles

Moles = mass / molar mass

Moles H2SO4 = 4.90 grams / 98.08 g/mol

Moles H2SO4 = 0.0500 moles

Moles lead acetate = 4.9 grams / 325.29 g/mol

Moles lead acetate = 0.0151 moles

Step 4: Calculate the limiting reactant

For 1 mol H2SO4 we need 1 mol lead acetate to produce 1 mol PbSO4 and 2 moles CH3COOH

The limiting reactant is lead acetate. It will completzly be consumed (0.0151 moles). H2SO4 is in excess. There will react 0.0151 moles. There will remain 0.0500 - 0.0151 = 0.0349 moles

Step 5: Calculate moles of products

For 1 mol H2SO4 we need 1 mol lead acetate to produce 1 mol PbSO4 and 2 moles CH3COOH

For 0.0151 moles lead acetate we'll have 0.0151 moles PbSO4 and 2*0.0151 = 0.0302 moles CH3COOH

Step 6: Calculate mass

Mass = moles * molar mass

Mass H2SO4 = 0.0349 moles * 98.08 g/mol

Mass H2SO4 = 3.42 grams

Mass PbSO4 = 0.0151 moles * 303.26 g/mol

Mass PbSO4 = 4.58 grams

Mass of CH3COOH = 0.0302 moles * 60.05 g/mol

Mass of CH3COOH = 1.81 grams

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Explanation:

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