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2 years ago

What information is needed to determine the orientation of an orbital?

Physics

1 answer:

kolbaska11 [484]2 years ago

8 0

Answer:

The magnetic quantum number (l) determines the orientation of an orbital

Explanation:

The magnetic quantum number of an electron's orbital is the spatial orientation of the electron's orbital

The magnetic quantum number, ml, specifies the orientation and number of orbitals of electrons in a subshell. The value of the magnetic quantum number is dependent on the angular momentum quantum number I with values ranging from -I to +I.

The shape of the electron's orbital is determined by the angular momentum quantum number.

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A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and

kari74 [83]

Answer:

$a. 4.2057\times 10^-^1^2 F \ or 4.2057\ pF\\b. 2.7344V$

Explanation:

Given the permittivity constant to be $8.854\times 10^-^1^2 F/m$ ,The capacitance of a $cylindrical \ capacitor$ of length, $L$ is given by the equation:

$C=\frac{2\pi \epsilon _o L}{ln(b/a)}$ where $b$ is the radius of the outer cylinder and $a$ the radius of the inner cylinder.

The values are given as: $a=0.550mm(5.5\times 10^-^4m), \ b=4.00mm(4.0\times10^-^3m), \ L=15.0cm(0.150m)$

Substitute in our capacitance equation:

$C=\frac{2\times\pi \times 8.854\times 10^-^1^2 \times 0.15}{In(4.00/0.550)}\\=4.2057\times 10^-^1^2 F$

Hence the capacitance is $4.2057\times 10^-^1^2 F$

b. The charge on the capacitance is related to the potential difference across it. The potential difference is expressed using the equation:

$Q=CV$ , $Q=11.5pC$

From a above, we already have our capacitance value, $C=4.2057\times 10^-^1^2 F$

We substitute $C$ in the pd equation:

$v=>(11.5)/(4.2057)\\=2.7344V$

Hence, the applied potential difference is 2.7344V

4 0

2 years ago

A storage tank containing oil (SG=0.92) is 10.0 meters high and 16.0 meters in diameter. The tank is closed, but the amount of o

blagie [28]

Answer:

a-1 Graph is attached. The relation is linear.

a-2 The corresponding height for 68 kPa Pressure is 7.54 m

a-3 The corresponding weight for 68 kPa Pressure is 1394726kg

b The original height of the column is 5.98 m

Explanation:

Part a

a-1

The graph is attached with the solution. The relation is linear as indicated by the line.

a-2

By the equation

$P=\rho \times g \times h$

Here

P is the pressure which is given as 68 kPa.
ρ is the density of the oil whose SG is 0.92. It is calculated as

$\rho=S.G \times \rho_{water}\\\rho=0.92 \times 1000 kg/m^3\\\rho=920 kg/m^3\\$

g is the gravitational constant whose value is 9.8 m/s^2
h is the height which is to be calculated

$P=\rho \times g \times h\\h=\frac{P}{\rho \times g}\\h=\frac{68 \times 10^3}{920 \times 9.8}\\h=7.54m$

So the height of column is 7.54m

a-3

By the relation of volume and density

$M=\rho \times V$

Here

ρ is the density of the oil which is 920 kg/m^3
V is the volume of cylinder with diameter 16m calculated as follows

$V=\pi r^2h\\V=3.14\times (8)^2 \times 7.54\\V=1515.23 m^3$

Mass is given as

$M=\rho \times V\\M=920 \times 1515.23\\M=1394726kg$

So the mass of oil leading to 68kPa is 1394726kg

Part b

Pressure variation is given as

$\Delta P=P_{obs}-P_{atm}\\\Delta P=115-101 kPa\\\Delta P=14 kPa\\$

Now corrected pressure is as

$P_c=P_g-\Delta P\\P_c=68-14 kPa\\P_c=54 kPa$

Finding the value of height for this corrected pressure as

$P_c=\rho \times g \times h\\h=\frac{P_c}{\rho \times g}\\h=\frac{54 \times 10^3}{920 \times 9.8}\\h=5.98m$

The original height of column is 5.98m

4 0

3 years ago

The current that charges a capacitor transfers energy that is stored in the capacitor’s electric field. Consider a 2.0 μF capaci

lapo4ka [179]

Answer:

the capacitor voltage is V = 20 V

Explanation:

Given,

Capacitance of the capacitor = 2.0 μF

energy stored = 200 W

time (t) =2.0 μs

Capacitor voltage = ?

$\dfrac{dE}{dt}=200 W$

$E =200\times 2 \times 10^{-6}$

$E =400 \times 10^{-6}\ J$

we know,

$E = \dfrac{1}{2}CV^2$

$V =\sqrt{\dfrac{2E}{C}}$

$V =\sqrt{\dfrac{2\times 400 \times 10^{-6}}{2\times 10^{-6}}}$

$V =\sqrt{400}$

V = 20 V

so , the capacitor voltage is V = 20 V

7 0

2 years ago

A light spring obeys Hooke's law. The spring's unstretched length is 34.0 cm. One end of the spring is attached to the top of a

sleet_krkn [62]

When the spring is extended by 44.5 cm - 34.0 cm = 10.5 cm = 0.105 m, it exerts a restoring force with magnitude R such that the net force on the mass is

∑ F = R - mg = 0

where mg = weight of the mass = (7.00 kg) g = 68.6 N.

It follows that R = 68.6 N, and by Hooke's law, the spring constant is k such that

k (0.105 m) = 68.6 N ⇒ k = (68.6 N) / (0.105 m) ≈ 653 N/m

5 0

1 year ago

If a chemical is completely consumed in a chemical reaction, what is the substance?

Dima020 [189]

Is a process in which One set of substances, called REACTANTS, is converted to a new set of substances is called PRODUCTS.

--In other words, a chemical reaction is the process by which a chemical change occurs.

5 0

3 years ago