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3 years ago

What is the maximum number of orbitals in the p sublevel?

Physics

1 answer:

dedylja [7]3 years ago

7 0

Answer:

3 orbitals

Explanation:

The p sublevel can carry a maximum of 6 electrons. Therefore it must be made of 3 orbitals which each, as in any orbital of any other sublevel, can carry 2 electrons. So there is a p sub x orbital, p sub y orbital and p sub z orbital which are all orthogonal to each other.

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Any help please, physics is not my strong suit?

frez [133]

Answer:

2000mg = 2g

5L = 5000mL

16cm = 160mm

5 0

3 years ago

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Below is an "oracle" function. An oracle function is a function presented interactively. When you type in a t value, and press t

Bingel [31]

Answer:

The rate of change of height with respect to time is -10.64 feet/sec

Explanation:

Given that,

There are three lines, so you can calculate three different values of the function at one time.

The function f(t) represents the height in feet of a ball thrown into the air, t seconds after it has been thrown.

Given table is,

Time t = 0, 1, 1,02

Function is, $F(t)=-3.053113177191196\times10^{-18},6.000000000000134, 6.41760000000015$

We need to calculate the initial height of ball

Using equation of motion

$h=h_{0}+ut-\dfrac{1}{2}gt^2$

Where, h₀ = initial height

u = initial velocity

t = time

g = acceleration due to gravity

At t = 0,

Put the value into the formula

$-3.053113177191196\times10^{-18}=h_{0}+0-0$

$h_{0}=-3.053113177191196\times10^{-18}$

We need to calculate the height of ball at t = 1

Using equation of motion

$h_{1}=h_{0}+u_{0}t-\dfrac{1}{2}gt^2$

Put the value in the equation

$6.000000000000134=-3.053113177191196\times10^{-18}+u-\dfrac{1}{2}\times32$

$6.000000000000134+3.053113177191196\times10^{-18}+16=u_{0}$

$u_{0}=22\ feet/s$

Velocity is the rate of change of height with respect to time

So, velocity at 1.02 sec is given

We need to calculate the height

Using equation of motion

$h=ut-\dfrac{1}{2}gt^2$

On differentiating w.r.to t

$h'(t)=u-\dfrac{1}{2}g(2t)$

Put the value into the formula

$h'(t)=22-\times32\times(1.02)$

$h'(t)=-10.64\ feet/sec$

Hence, The rate of change of height with respect to time is -10.64 feet/sec.

4 0

3 years ago

A concave mirror has a radius of curvature of 1.6m. Find the focal length

Yuki888 [10]

Given,

Radius of curvature of concave mirror = 1.6m

We know that ,

Focal length = radius/2

Hence ,

Focal length of concave mirror = radius of concave mirror /2

=> F = 1.6/2

=> F = 0.8m

Hence the focal length of concave mirror is 0.8 m

5 0

2 years ago

Read 2 more answers

Will you still get tan with sunscreen on

WARRIOR [948]

No, wearing sunscreen won't let you tan. The whole purpose of sunscreen is to protect your skin from sun rays. You can start to become tan if you dont have much sunscreen on - its happened to me.

5 0

4 years ago

An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com

solong [7]

Answer:

(a) $3.81\times 10^5\ Pa$

(b) $4.19\times 1065\ Pa$

Explanation:

<u>Given:</u>

$T_1$ = The first temperature of air inside the tire = $10^\circ C =(273+10)\ K =283\ K$

$T_2$ = The second temperature of air inside the tire = $46^\circ C =(273+46)\ K= 319\ K$

$T_3$ = The third temperature of air inside the tire = $85^\circ C =(273+85)\ K=358 \ K$

$V_1$ = The first volume of air inside the tire

$V_2$ = The second volume of air inside the tire = $30\% V_1 = 0.3V_1$

$V_3$ = The third volume of air inside the tire = $2\%V_2+V_2= 102\%V_2=1.02V_2$

$P_1$ = The first pressure of air inside the tire = $1.01325\times 10^5\ Pa$

<u>Assume:</u>

$P_2$ = The second pressure of air inside the tire

$P_3$ = The third pressure of air inside the tire
n = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

$PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)$

Part (a):

Using the above equation for this part of compression in the air, we have

$\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa$

Hence, the pressure in the tire after the compression is $3.81\times 10^5\ Pa$ .

Part (b):

Again using the equation for this part for the air, we have

$\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa$

Hence, the pressure in the tire after the car i driven at high speed is $4.19\times 10^5\ Pa$ .

8 0

3 years ago