All categories

3 years ago

What are some guidelines for using self-assessments?

2 answers:

4 0

Decide which components you wish to assess yourself. There is no set order in which you do this. you could, for an example assess one or two elements and skip the rest.

Dmitriy789 [7]3 years ago

3 0

Answer:

Few guidelines which are choosing for the personal self assessments, that work for your best and raise your health standards, instead of others doing this on the behalf of yours.

Explanation:

There are some guidelines which are discussed below for using self assessments.

The observer can learn to organize his learning autonomously.
The observer can motivate himself in that way that will work for him.
The observer should be aware of his motivational learning and can rely on that.
He can re motivate himself again after knowing that his initial motivation is growing thin.
He can motivate himself to learn something which are good for his personal growth and development.

You might be interested in

A charge q1 of -5.00 x 10^-9 C and a charge q2 of -2.00x 10^-9 C are separated by a distance of 40.0 cm. Find the equilibrium po

Blababa [14]

The magnitude of charge on a proton and electron is the same, 1.602 x 10-19 C. Protons are +, and electrons -.

5 0

3 years ago

A long wire carrying a 5.0 A current perpendicular to the xy-plane intersects the x-axis at x= - 2.0 cm . A second, parallel wir

mario62 [17]

Answer:

a . 0.35cm

b. 11.33cm

Explanation:

a. Given both currents are in the same direction, the null point lies in between them. Let x be distance of N from first wire, then distance from 2nd wire is 4-x

#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in between the wires:

$\frac{\mu_oi_1}{2\pi x}=\frac{\mu_oi_2}{2\pi(4-x)}\\\\5/x=\frac{3.5}{4-x}\\\\x=2.35cm\\\\N=2.35-2=0.35cm$

Hence, for currents in same direction, the point is 0.35cm

b. Given both currents flow in opposite directions, the null point lies on the other side.

#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in outside the wires:

Let x be distance of N from first wire, then distance from 2nd wire is 4+x:

$\frac{\mu_oi_1}{2\pi(4+ x)}=\frac{\mu_oi_2}{2\pi x}\\\\5/(4+x)=\frac{3.5}{x}\\\\x=9.33cm\\\\N=9.33+2=11.33cm$

Hence, if currents are in opposite directions the point on x-axis is 11.33cm

8 0

3 years ago

Three capacitors having capacitances of 8.40, 8.40, and 4.20 μFμF, respectively, are connected in series across a 36.0-V potenti

son4ous [18]

Answer:

a)Q=71.4 μ C

b)ΔV' = 10.2 V

Explanation:

Given that

C ₁= 8.7 μF

C₂ = 8.2 μF

C₃ = 4.1 μF

The potential difference of the battery, ΔV= 34 V

When connected in series

1/C = 1/C ₁ + 1/C₂ + 1/C₃

1/ C= 1/8.4 +1 / 8.4 + 1/4.2

C=2.1 μF

As we know that when capacitor are connected in series then they have same charge,Q

Q= C ΔV

Q= 2.1 x 34 μ C

Q=71.4 μ C

As we know that when capacitor are connected in parallel then they have same voltage difference.

Q'= C' ΔV'

C'= C ₁+C₂+C₃ (For parallel connection)

C'= 8.4 + 8.4 + 4.2 μF

C'=21 μF

Q'= C' ΔV'

Q'=3 Q

3 x 71.4= 21 ΔV'

ΔV' = 10.2 V

3 0

3 years ago

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

kristine speeds past a parked police car at 32 m/s. The police car starts from rest with a uniform acceleration of 2.5 m/s^2. Ho

Digiron [165]

$32 = 0 \times t + \frac{1}{2} \times 2.5 \times t^{2} \\ 32 = 0 + 1.25 \times t {}^{2} \\ 32 = 1.25t {}^{2} \\ \frac{32}{1.25} = \frac{1.25t {}^{2} }{1.25} \\ t {}^{2} = 25.6 \\ \sqrt{t {}^{2} } = \sqrt{25.6} \\ t = 5.1seconds \\$

7 0

3 years ago