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Sauron [17]
3 years ago
12

What is the maximum angular momentum Lmax that an electron with principal quantum number n = 2 can have? Express your answer in

units of ℏ.
Physics
1 answer:
butalik [34]3 years ago
4 0

Answer:

L_{max} = 1.414 ℏ

Given:

Principle quantum number, n = 2

Solution:

To calculate the maximum angular momentum, L_{max}, we have:

L_{max} = \sqrt {l(1 + l)}                              (1)

where,

l = azimuthal quantum number or angular momentum quantum number

Also,

n = 1 + l

2 = 1 + l

l = 1

Now,

Using the value of l = 1 in eqn (1), we get:

L_{max} = \sqrt {1(1 + 1)} = \sqrt 2

L_{max} = 1.414 ℏ

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1 poir
djverab [1.8K]

Answer:

D) 735 J(oules)

Explanation:

Work is defined as force * distance

Force is defined as mass * acceleration

Given a mass of 15 kg and a gravitational acceleration of 9.8 m/s² since the box is being lifted up, the force being applied to the box is 15 kg * 9.8 m/s² = 147 N

Since the distance is 5 meters, the work done is 147 N * 5 m = 735 N/m = 735 J, making D the correct answer.

4 0
3 years ago
vector u has a magnitude of 20 and direction of 0°.vector v has amagnitude of 40and a direction of 60°.find the magnitude and di
pantera1 [17]

Addition of vectors:

vector u

has a magnitude of 20 and a direction of 0º with respect to the horizontal, vector v has a magnitude of 40 and a direction of 60º with respect to the horizontal.

a) Find the magnitude and direction of the resultant to the nearest whole

number.

Vector Sum:

The resultant of two vectors is simply the vector sum of the vectors. There are a handful of ways to present the resultant factor; the notation that shows the vector magnitude and direction is called the polar vector notation. An example of a vector presented in polar vector notation is

a∠θ where a is the magnitude and θis the angle that the vector makes with the horizontal axis.

Answer and Explanation:

Let's first present the vectors in rectangular vector notation.

For the vector →u of magnitude 20 and direction 0∘ to the horizontal axis, the vector is →u=^i20.

For the vector →v

of magnitude 40 and direction 60∘ to the horizontal axis, the vector is →v=^i40cos60∘+^j40sin60∘.

The resultant vector →w is the vector sum of the vectors, i.e.

→w=→u+→v

=^i20+^i40cos60∘+^j40sin60∘

=^i(20+40cos60∘)+^j(40sin60∘)

=^i40+^j20√3

For a vector ^ix+^jy, the magnitude of the vector is √x2+y2 and the direction above the horizontal axis is θ=tan−1(yx).

Let's use the formulas:

|→w|∠θ=√(40)2+(20√3)2∠tan−1(20√340)≈52.9∠40.9∘

The magnitude of the vector is about 53 units in the direction 41-degrees above the horizontal axis.

6 0
3 years ago
An earthquake’s epicenter is _____.
Evgen [1.6K]
The point in which it originates.
4 0
3 years ago
Read 2 more answers
Two power lines run parallel for a distance of 222 m and are separated by a distance of 40.0 cm. if the current in each of the t
earnstyle [38]
1) Magnitude of the force:

The magnetic field generated by a current-carrying wire is
B= \frac{\mu_0I}{2 \pi r}
where
\mu_0 is the vacuum permeability
I is the current in the wire
r is the distance at which the field is calculated

Using I=135 A, the current flowing in each wire, we can calculate the magnetic field generated by each wire at distance 
r=40.0 cm=0.40 m, 
which is the distance at which the other wire is located:
B= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} N/A^2)(135 A) }{2 \pi (0.40 m)}=6.75 \cdot 10^{-5} T

Then we can calculate the magnitude of the force exerted on each wire by this magnetic field, which is given by:
F=ILB=(135 A)(222 m)(6.75 \cdot 10^{-5}T)=2.03 N

2) direction of the force: 
The two currents run in opposite direction: this means that the force between them is repulsive. This can be determined by using the right hand rule. Let's apply it to one of the two wires, assuming they are in the horizontal plane, and assuming that the current in the wire on the right is directed northwards:
- the magnetic field produced by the wire on the left at the location of the wire on the right is directed upward (the thumb of the right hand is directed as the current, due south, and the other fingers give the direction of the magnetic field, upward)

Now let's apply the right-hand rule to the wire on the right:
- index finger: current --> northward
- middle finger: magnetic field --> upward
- thumb: force --> due east --> so the force is repulsive

A similar procedure can be used on the wire on the left, finding that the force exerted on it is directed westwards, so the force between the two wires is repulsive.
6 0
3 years ago
PLEASE help me quickly
Anna007 [38]

Explanation:

its either a or d however i say the best choice is d

6 0
3 years ago
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