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inessss [21]
3 years ago
11

What is the magnitude of the gravitational field midway between the earth and moon? ignore effects of the sun.

Physics
1 answer:
bazaltina [42]3 years ago
3 0
As it is considered the gravitational field is zero at the point where the gravitational pull of the moon equals that of the earth but it is not at the middway point
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5 0
4 years ago
Light travels at a speed of 3.00 x 10^11 cm/s. What is the speed of light in kilometers/hour?
rewona [7]

As 1 km = 1000 m = 1000,00 cm,

So, 1 cm = (1/1000,00) km

1 hour = 60 × 60 s = 3600 s

So,  1 s = (1 / 3600) hour

The light travels at a speed of 3.00 \times 10^{11}  \ cm/s.

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4 0
4 years ago
Find the minimum value of n in the Balmer series for which the predicted wavelength is in the ultraviolet region of the spectrum
d1i1m1o1n [39]

Answer:

 λ =365.4 nm

Explanation:

Boh's atomic model of the Hydrogen atom the energy of each level is

        Eₙ = - 13.606 / n²

where the synergy is in electonvotes and the value of E₀ = 13.606 eV is the energy of the base state of hydrogen.

An atomic transition occurs when an electron goes from an excited state and joins everything of lower energy.

                 ED = 13.606 (1 / n₀² - 1 /n_{f}^{2})

we are going to apply this relationship to answer slash.

 

At the beginning of the studies of atomic transitions, each group did not consider having a different name

name        Initial state

Lymman         1

Balmer           2

the final state is any other state sta the continuum that corresponds to n = inf

Let's look for the highest energy of the Balmer series

              ΔE = 13.606 (1/2² - 1 /∞)

              ΔE = 3.4015 eV

Let's use the Planck relation for the energy

                E = h f = h c /λ

                λ = h c / E

Let's reduce the energy to J

              E = 3.4015 eV (1.6 10⁻¹⁹ J / 1 eV) = 5.4424 10⁻¹⁹

            λ = 6.63 10⁻³⁴  3 10⁸ / 5.4424 10⁻¹⁹

            λ = 3.654 10⁻⁷ m

            λ = 3,654 10⁻⁷ m (10⁹ nm / 1m)

            λ =365.4 nm

this eta radiation in the ultraviolet range

5 0
3 years ago
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