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ohaa [14]
4 years ago
10

If root-mean-square voltage of a supply is 240 V, calculate the maximum voltage of

Physics
1 answer:
Levart [38]4 years ago
5 0

Answer:

Vp= Vr sq. root 2

= 240 × sq. root 2

=339.4 V

You might be interested in
Percent Yield Lab Report
Vlada [557]

Based on the data obtained from the reaction, the following conclusions can be made;

  • according to the law of conservation of mass, the mass of the product was higher than the reactant because of the mass of oxygen added during the combustion.
  • the percent yield is less than 100% because of loss in mass of either reactants or products.
  • the errors could have occurred during the weighing and transfer of reactants and products
  • repeated measurements are required in order to improve accuracy

<h3>What is the percent yield of the reaction?</h3>

Equation of the reaction is given below:

  • 2 Mg + O₂ ----> 2 MgO

Trial 1

Mass of empty crucible with lid = 26.698 (g)

Mass of Mg metal, crucible, and lid  = 27.040 (g)

Mass of MgO, crucible, and lid = 27.198 (g)

Mass of metal = 27.040 - 26.698 = 0.342

Mass of MgO = 27.198 - 26.698 = 0.500

<h3>Moles of Mg used</h3>

moles of Mg = mass/molar mass

  • molar mass of Mg = 24 g

moles of Mg = 0.342/24 = 0.01425

<h3>Moles of MgO expected</h3>

Based on the equation of reaction;

moles of MgO expected = 0.01425

moles of MgO produced =  mass/molar mass

  • molar mass of MgO = 40 g/mol

moles of MgO produced = 0.500/40 = 0.0125

<h3>Percent yield</h3>
  • Percent yield = (moles of MgO produced/moles of MgO expected) * 100%

Percentage yield = 0.0125/0.01425 * 100%

Percent yield of MgO = 87.7%

Trial 2

Mass of empty crucible with lid = 26.691 (g)

Mass of Mg metal, crucible, and lid = 27.099 (g)

Mass of MgO, crucible, and lid = 27.361 (g)

Mass of metal = 27.099 - 26.691 = 0.408

Mass of MgO = 27.361 - 26.691 = 0.670

<h3>Moles of Mg used</h3>

moles of Mg = mass/molar mass

  • molar mass of Mg = 24 g

moles of Mg = 0.408/24 = 0.0170

<h3>Moles of MgO expected</h3>

Based on the equation of reaction;

moles of MgO expected = 0.0170

  • moles of MgO produced =  mass/molar mass

molar mass of MgO = 40 g/mol

moles of MgO produced = 0.670/40 = 0.01675

<h3>Percent yield</h3>
  • Percent yield = (moles of MgO produced/moles of MgO expected) * 100%

Percent yield = 0.01675/0.0170 * 100%

Percent yield of MgO = 98.5%

Average percent yield = (87.7 + 98.5)% / 2

Average percent yield = 89.0%

Based on the data obtained from the reaction, the following conclusion can be made;

  • according to the law of conservation of mass, the mass of the product was higher than the reactant because of the mass of oxygen added during the combustion.
  • the percent yield is less than 100% because of loss in mass of either reactants or products.
  • the errors could have occurred during the weighing and transfer of reactants and products
  • repeated measurements are required in order to improve accuracy

Learn more about law of conservation of mass at: brainly.com/question/1824546

6 0
3 years ago
I’m in the middle of a test right now. If anybody knows this can they help
babymother [125]

Answer:

C

Explanation:

momentum = mass ×velocity

A. mv = 200

B. mv = 300

C. mv= 400

D. mv= 200

highest is C.

6 0
3 years ago
A hobby rocket reaches a height of 72.3 m and lands 111 m from the launch point with no air resistance. What was the angle of la
Deffense [45]

Answer:

The angle of launch is 52.49 Degree.

Explanation:

The Range R and Height H of a thrown object is calculated using the formula,

R=V₀² sin(2φ)/g

H=V₀²sin²(φ)/g

From these equations it can be written,

V₀²=R g/ sin(2φ)

V₀²=H g/ sin²(φ)

These values are equal so it can be written by equating these equations,

R g/sin(2φ)=H g/sin²(φ)

tan(φ)= 2H/R

Given H=72.3 m and R=111 m, the angle of launch is,

tan(φ)= 2*72.3/111

φ= 52.49 Degree.

Check out other solutions,

brainly.com/question/1495042

#SPJ10

7 0
2 years ago
This table shows the weights of four different objects that are sitting on the
bekas [8.4K]
I think that it would be A
6 0
3 years ago
What’s the mathematical relation between them
Gekata [30.6K]

Explanation:

there is no relationship between small mass and the bigger mass, but it can be related with the acceleration. Since Force is constant, acceleration is inversely proportional to the mass. Greater the mass, lesser is the acceleration and vise versa

6 0
3 years ago
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