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nataly862011 [7]
2 years ago
15

The rate of change of momentum of a body free falling under gravity is equal to its? A. Velocity B. kinetic energy C. power D. w

eight
Physics
2 answers:
jolli1 [7]2 years ago
7 0

Answer:

weight

Explanation:

this is because of the lighter the object the faster it will be affected

ludmilkaskok [199]2 years ago
4 0
Your correct answer would be weight.
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Answer:

A. 2.82 eV

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C. 59.5 angstroms

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E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})     (1)

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n2: initial state = 2

Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV

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E=h\frac{c}{\lambda}   (2)

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c: speed of light = 3*10^8 m/s

λ: wavelength of the photon

You first convert the energy from eV to J:

2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J

Next, you use the equation (2) and solve for λ:

\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm

C. The radius of the orbit is given by:

r_n=n^2a_o   (3)

where ao is the Bohr's radius = 2.380 Angstroms

You use the equation (3) with n=5:

r_5=5^2(2.380)=59.5

hence, the radius of the atom in its 5-th state is 59.5 anstrongs

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