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Lana71 [14]
4 years ago
14

How many electrons will fit on an outermost shell of an atom?

Physics
1 answer:
Kaylis [27]4 years ago
7 0
B. 8 is the answer but sometimes its determined by the quantum numbers on the chart

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Which graph shows the relationship between temperature, X, and kinetic energy, Y?
strojnjashka [21]
Answer: see the graph attached (straight line, passing through the origin and positive slope).


Justification:

1) Kinetic energy and temperature are in direct proportion. That means:

i)  Being kinetic energy y and temperature x: y α x

ii) That implies: y = kx,where k is the constant of proportionality.

iii) The graph is a line that passes through the origin and has positive slope k (k = y / x).

2) The proportional relationship between kinetic energy (KE) and temperature (T) is shown by the Boltzman law, which states:

Average KE = [3 / 2] KT, where K is Boltzman's constant, whose graph is of the form shown in the figure attached.

5 0
4 years ago
Read 2 more answers
What are two examples of common units for each of the above measurements
kap26 [50]

Density: g/mL, kg/cubic meter  

Volume: L, teaspoon  

Mass: g, MeV/sq. C

3 0
4 years ago
f your risk-aversion coefficient is A = 4 and you believe that the entire 1926–2015 period is representative of future expected
siniylev [52]

Answer:

The portfolio should invest 48.94% in equity while 51.05% in the T-bills.

Explanation:

As the complete question is not given here ,the table of data is missing which is as attached herewith.

From the maximized equation of the utility function it is evident that

Weight=\frac{E_M-r_f}{A\sigma_M^2}

For the equity, here as

  • Weight is percentage of the equity which is to be calculated
  • {E_M-r_f} is the Risk premium whose value as seen from the attached data for the period 1926-2015 is 8.30%
  • A is the risk aversion factor which is given as 4.
  • \sigma_M is the standard deviation of the portfolio which from the data for the period 1926-2015 is 20.59

By substituting values.

Weight=\frac{E_M-r_f}{A\sigma_M^2}\\Weight=\frac{8.30\%}{4(20.59\%)^2}\\Weight=0.4894 =48.94\%

So the weight of equity is 48.94%.

Now the weight of T bills is given as

Weight_{T-Bills}=1-Weight_{equity}\\Weight_{T-Bills}=1-0.4894\\Weight_{T-Bills}=0.5105=51.05\%\\

So  the weight of T-bills is 51.05%.

The portfolio should invest 48.94% in equity while 51.05% in the T-bills.

7 0
3 years ago
3. A 2kg wooden block whose initial speed is 3 m/s slides on a smooth floor for 2 meters before it comes to a
serious [3.7K]

Answer:

Calculating Coefficient of friction is 0.229.

Force is 4.5 N that keep the block moving at a constant speed.

Explanation:

We know that speed expression is as \mathrm{V}^{2}=\mathrm{V}_{\mathrm{i}}^{2}+2 . \mathrm{a} . \Delta \mathrm{s}.

Where, {V}_{i} is initial speed, V is final speed, ∆s displacement and a acceleration.

Given that,

{V}_{i} =3 m/s, V = 0 m/s, and  ∆s = 2 m

Substitute the values in the above formula,

0=3^{2}-2 \times 2 \times a

0 = 9 - 4a

4a = 9

a=2.25 \mathrm{m} / \mathrm{s}^{2}

a=2.25 \mathrm{m} / \mathrm{s}^{2} is the acceleration.

Calculating Coefficient of friction:

\mathrm{F}=\mathrm{m} \times \mathrm{a}

\mathrm{F}=\mu \times \mathrm{m} \times \mathrm{g}

Compare the above equation

\mu \times m \times g=m \times a

Cancel "m" common term in both L.H.S and R.H.S

\text { Equation becomes, } \mu \times g=a

\text { Coefficient of friction } \mu=\frac{a}{g}

\mathrm{g} \text { on earth surface }=9.8 \mathrm{m} / \mathrm{s}^{2}

\mu=\frac{2.25}{9.8}

\mu=0.229

Hence coefficient of friction is 0.229.

calculating force:

\text { We know that } \mathrm{F}=\mathrm{m} \times \mathrm{a}

\mathrm{F}=2 \times 2.25 \quad(\mathrm{m}=2 \mathrm{kg} \text { given })

F = 4.5 N

Therefore, the force would be <u>4.5 N</u> to keep the block moving at a constant speed across the floor.

7 0
3 years ago
When an electric stove element is hot enough, it gives off a dull red glow. When it cools to the point that it no longer glows,
DochEvi [55]

Answer:

It will have a longer wavelength

Explanation:

When an electric stove is hot and gives dull red glow. a part of the energy dissipated is emitted as visible light and part as infrared radiation in the form of heat. When the stove cools down, and no longer glows all the energy is now in the form of infrared radiation.In the electromagnetic spectrum infrared rays have a higher wavelength than visible light. Hence for the reason the radiation will have a higher wavelength since visible light is cut off.

4 0
3 years ago
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