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3 years ago

8

the Space Program has benefitted people in everyday life. Describe two ways in which people can benefit

1 answer:

ycow [4]3 years ago

8 0

So we can know what is in space maybe weird or interesting stuff

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A cd player uses a ___ to convert info to an electrical signal

Alex_Xolod [135]

C.) Laser. the light from the laser reflects off the shiny surface as the CD rotates

7 0

3 years ago

A 1900kg car starts from rest and drives around a flat 65-m-diameter circular track. The forward force provided by the car's dri

s344n2d4d5 [400]

Answer:

$The\quad magnitude\quad of\quad the\quad car's\quad acceleration\quad at\quad t=13s\quad \quad =2.52m/{ s }^{ 2 }\\ The\quad direction\quad of\quad the\quad car's\quad acceleration\quad at\quad t=13s\quad =15.{ 72 }^{\o}\\The\quad car\quad begins\quad to\quad slide\quad out\quad \quad of\quad the\quad circle\quad after\quad 26.09s.\quad \quad \quad \quad$

Explanation:

8 0

3 years ago

Angelina_Jolie [31]

I am very sorry I don’t know

7 0

3 years ago

If you ride your bicycle down a straight road for 500m then turn around and ride back, your distance is____ your displacement.

Helen [10]

Answer:

C-less than

Explanation:

Distance is total distance traveled (1000m here if you stop where you started).

Displacement is your final distance from where you started (0m if you stop where you started).

0m<1000m

4 0

3 years ago

Read 2 more answers

Use the Bohr model to address this question. When a hydrogen atom makes a transition from the 5 th energy level to the 2nd, coun

iris [78.8K]

Answer:

A. 2.82 eV

B. 439nm

C. 59.5 angstroms

Explanation:

A. To calculate the energy of the photon emitted you use the following formula:

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

n1: final state = 5

n2: initial state = 2

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

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

B. The energy of the emitted photon is given by the following formula:

$E=h\frac{c}{\lambda}$ (2)

h: Planck's constant = 6.62*10^{-34} kgm^2/s

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

8 0

3 years ago