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Shkiper50 [21]
3 years ago
14

A north magnetic pole is facing another north magnetic pole with a distance x. If the distance between the poles becomes 12x, wh

at happens to the magnitude of the field energy between them?
The answer is "The field energy will increase"
Physics
1 answer:
krek1111 [17]3 years ago
4 0

Answer:

1.)The field energy will increase.

The rest of the answers:

2.)The energy increases, and the lines of force are denser

3.) It points toward the field of earths magnetic poles.

4.) l, ll, and lll only

5.) ll, lV, l, lll

You might be interested in
An electron is trapped in an infinite square-well potential of width 0.6 nm. If the electron is initially in the n = 4 state, wh
grandymaker [24]

Answer:

E₁ = 1.042 eV

E₄₋₃= 7.29 eV      

E₄₋₂= 12.50 eV

E₄₋₁= 15.63 eV

E₃₋₂= 5.21eV

E₃₋₁= 8.34eV

E₂₋₁= 3.13eV

Explanation:

The energy in an infinite square-well potential is giving by:  

E = \frac {h^{2} n^{2}}{8mL^{2}}      

<em>where, h: Planck constant = 6.62x10⁻³⁴J.s, n: is the energy state, m: mass of the electron and L: widht of the square-well potential </em>      

<u>The energy of the electron in the ground state, </u><u>n = 1</u><u>, is:  </u>

E_{1} = \frac {(6.62 \cdot 10^{-34})^{2} (1)^{2}}{(8) (9.11 \cdot 10^{-31}) (0.6 \cdot 10^{-9} m)^{2}}    

E_{1} = 1.67 \cdot 10^{-19} J = 1.042 eV      

The photon energies that are emitted as the electron jumps to the ground state is the difference between the states:                      

E_{\Delta n} = \Delta n^{2} E_{1}  

E_{(4 - 3)} = (4^{2} - 3^{2}) 1.042 eV = 7.29eV

E_{(4 - 2)} = (4^{2} - 2^{2}) 1.042 eV = 12.50eV

E_{(4 - 1)} = (4^{2} - 1^{2}) 1.042 eV = 15.63eV

E_{(3 - 2)} = (3^{2} - 2^{2}) 1.042 eV = 5.21eV

E_{(3 - 1)} = (3^{2} - 1^{2}) 1.042 eV = 8.34eV

E_{(2 - 1)} = (2^{2} - 1^{2}) 1.042 eV = 3.13eV    

Have a nice day!                          

7 0
2 years ago
An archer pulls a bow string 0.5 m. If the spring constant is 16,000 N/m, what is the energy stored in the bow string?
mars1129 [50]

2000J

Explanation:

Given parameters:

Extension = 0.5m

Spring constant = 16000N/m

Unknown:

Energy stored in the bow string = ?

Solution:

The energy stored in a bow string is an elastic potential energy.

It can be calculated using the expression below;

     Elastic energy = \frac{1}{2} K e²

Where k is the spring constant

            e is the extension

Input the parameters;

  Elastic energy = \frac{1}{2} K e²

                          =\frac{1}{2} x 16000 x 0.5²

                          = 2000J

learn more:

Potential energy brainly.com/question/10770261

#learnwithBrainly

3 0
3 years ago
How would you find the horizontal net force for the free body diagram below
tiny-mole [99]

Answer:

Add Ff from Fa

Explanation:

Fnet = sum of all force

horizontal net force = Ff + Fa

7 0
2 years ago
A 60-kg swimmer suddenly dives horizontally from a 150-kg raft with a speed of 1.5 m/s. The raft is initially at rest. What is t
scZoUnD [109]
I’m almost positive it 60 m/s
7 0
2 years ago
Formation of hydrogen bonds requires hydrogen atoms and what else?
ohaa [14]

Answer:

either a Nitrogen atom, Oxygen atom, or a Flourine atom

Explanation:

The atom has to be more electronegative than hydrogen for the bond to form.

8 0
3 years ago
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