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melamori03 [73]
3 years ago
11

imagine that you are on a spaceship that is far from the sun or any planet. Explain why a 500 kg piece of equipment would be dif

ficult to move, even though the object would be weightless?
Physics
1 answer:
Mila [183]3 years ago
8 0

Even though the object is weightless, it would need inertia, I.e, you pushing it or any form of transportation. So you would still have to push that 500kg to just keep it moving in space. Say if it were a planet with less gravitational force, it would be weightless.

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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
3 years ago
When you double the distance between a pair of charged particles what happens to the force between them?
Volgvan

Answer:

Force between the two charges becomes one fourth of the initial force.

Explanation:

The electrostatic force acting between any two charges is given as,

F = k\frac{q_{1}q_{2}}{r^{2}}

Here,

F = force

k = Coulomb's constant

q_{1} = magnitude of charge of the first particle

q_{2} = magnitude of charge of the second particle

r = separation between the two charges

From the above relation,

F \propto \frac{1}{r^{2}}

Thus,

\frac{F_{1}}{F_{2}} = \left ( \frac{r_{2}}{r_{1}} \right )^{2}

\frac{F_{1}}{F_{2}} = \left ( \frac{2r}{r} \right )^{2}

\Rightarrow \ F_{2} = \frac{1}{4}F_{4}.

8 0
3 years ago
Until a train is a safe distance from the station, it must travel at 5 m/s. Once ti
LuckyWell [14K]

Answer:

60

Explanation:

3 0
3 years ago
Pleaseeee help I really need to pass to graduate
Rus_ich [418]

Answer:

kWh I think

Explanation:

7 0
3 years ago
What is the name given to a material with zero resistance that can conduct electricity without a loss of energy?
tatyana61 [14]
It would be C. Superconductor 
3 0
3 years ago
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