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Anna [14]
3 years ago
13

Give a combination of four quantum numbers that could be assigned to an electron occupying a 5p orbital.

Physics
1 answer:
Elden [556K]3 years ago
8 0

Answer:

n=5, l=1, m(l) = -1, m(s)= + 1/2

Explanation:

Quantum number are used to describe the position and spin of an electron inside an atom. There are four types of quantum number for describing an electron inside an atom. They are: the principal quantum number, spin quantum number, magnetic quantum number and angular momentum quantum number.

(1).PRINCIPAL QUANTUM NUMBER: denoted by n, and has possible values of n= 1,2,3,4,.... IN HERE, n= 5

(2).ANGULAR MOMENTUM QUANTUM NUMBER: it is denoted by l, and has possible values of l= 0,1,2,3,...,(n-1).

Our l here is one( that is, s-orbital=0, p-orbital=1, d-orbital= 3 and so on)

(3).MAGNETIC QUANTUM NUMBER: The magnetic quantum number, which is denoted by m subscribt l, specifies the exact orbital in which you can find the electron. It has values ranging from -l,...,-1,0,1,...,l.

Here, our value is -1 that is m(l)= -1

(4).SPIN QUANTUM NUMBER: describes the orientation of electrons. Electrons can only have two values here, either a positive one and the half(+1/2) that is the spin up electron or the negative one and half(-1/2) that is the spin down electron.

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Convert 3402kgm/s to 20000Newtons
oee [108]

The 3,402 has units of kg-m/s.  That's momentum.  The 20,000 has units of Newtons.  That's force.  Momentum and force are different physical things, and you can't convert them from one to the other.

The best I can do for you is something like this:

Let's say you have a moving object with 3,402 kg-m/s of momentum, and you want to STOP it completely.  You want to stand in front of it and push back on it, hard enough and for long enough to CHANGE its momentum from 3,402 kg-m/s to zero.

Also ... there's a limit to how hard you can push.  The most force you can exert is 20,000 Newtons.

The amount you'll change its momentum is called the <u><em>impulse</em></u> you give it.  The quantity of impulse is (force) x (length of time you push on it).

So you need to keep pushing it back for (T seconds) long enough so that

(20,000 Newtons of force) x (T seconds) = 3,402 kg-m/s of momentum .

Divide each side of that equation by (20,000 Newtons). Then it says:

(T seconds) = (3,402 kg-m/s) / (20,000 Newtons)

<em>T = 0.1701 second</em>

And that's how you provide just enough impulse to stop the flying object ... push on it with 20,000 Newtons of force for exactly 0.1701 second, and it loses all its momentum and falls out of the air onto the ground at your feet.

This story is the closest I can come to anything that looks like "convert"ing momentum into force.

3 0
3 years ago
A bicyclist travels in a circle of radius 30.0 m at a constant speed of 8.00 m/s. the bicycle-rider mass is 82.0 kg calculate ne
IceJOKER [234]
F=m*(v^2/r)
F=82*(8^2/30)
F=174.9N
7 0
3 years ago
A flat coil of wire consisting of 15 turns, each with an area of 40 cm 2, is positioned perpendicularly to a uniform magnetic fi
zheka24 [161]

Answer:

0.54 A

Explanation:

Parameters given:

Number of turns, N = 15

Area of coil, A = 40 cm² = 0.004 m²

Change in magnetic field, ΔB = 5.1 - 1.5 = 3.6 T

Time interval, Δt = 2 secs

Resistance of the coil, R = 0.2 ohms

To get the magnitude of the current, we have to first find the magnitude of the EMF induced in the coil:

|V| = |(-N * ΔB * A) /Δt)

|V| = | (-15 * 3.6 * 0.004) / 2 |

|V| = 0.108 V

According to Ohm's law:

|V| = |I| * R

|I| = |V| / R

|I| = 0.108 / 0.2

|I| = 0.54 A

The magnitude of the current in the coil of wire is 0.54 A

6 0
3 years ago
This picture represents<br> A: Reflection<br> B: Refraction<br> C: Interference<br> D: Diffraction
-BARSIC- [3]

Answer:

I think it's a because it goes thru it and reflects

5 0
2 years ago
17 copper wires of length l and diameter d are connected in parallel to form a single composite conductor of resistance R. What
Lubov Fominskaja [6]

Answer:

\frac{D}{d} = 4.12

Explanation:

As we know that resistance of one copper wire is given as

r = \rho \frac{L}{a}

here we know that

a = \pi (\frac{d}{2})^2

now we have

r = \rho \frac{L}{\pi (\frac{d^2}{4})}

r = \rho \frac{4L}{\pi d^2}

now we know that such 17 resistors are connected in parallel so we have

R = \frac{r}{17}

R = \rho \frac{4L}{17 \pi d^2}

Now if a single copper wire has same resistance then its diameter is D and it is given as

R = \rho \frac{4L}{\pi D^2}

now from above two equations we have

\rho \frac{4L}{\pi D^2} = \rho \frac{4L}{17 \pi d^2}

D^2 = 17 d^2

now we have

\frac{D}{d} = 4.12

3 0
3 years ago
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