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Makovka662 [10]
3 years ago
9

Assuming there are no accidents or delays, the distance that a car travels down the interstate

Physics
1 answer:
Ganezh [65]3 years ago
7 0

Explanation:

The distance that a car travels down the interstate  can be calculated with the following formula:

Distance = Speed x Time

(A) Speed of the car, v = 70 miles per hour = 31.29 m/s

Time, d = 6 hours = 21600 s

Distance = Speed x Time

D = 31.29 m/s × 21600 s

D = 675864 meters

or

D=6.75\times 10^5\ m

(b) Time, d = 10 hours = 36000 s

Distance = Speed x Time

D = 31.29 m/s × 36000 s

D = 1126440 meters

or

D=1.12\times 10^6\ m

(c) Time, d = 15 hours = 54000 s

Distance = Speed x Time

D = 31.29 m/s × 54000 s

D = 1689660 meters

or

D=1.68\times 10^6\ m

Hence, this is the required solution.

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Please Help !!!!!!!!!!!!!!!!!!!!!!!!!!
larisa [96]

As per the question the wavelength of spectral line of the neutral hydrogen atom is 21 cm.

Spectral line of hydrogen atom will be observed when an electron will jump from higher excited to the ground state. If E is the energy of the ground state and E' is the energy of the higher excited state,the energy emitted when electron will jump from higher excited to ground state is E'-E.

As per Planck's quantum theory -

                                                      hf =E'-E

          Where f is the frequency of emitted radiation and h is the Planck's constant.Actually this energy emitted is also electromagnetic in nature. We know that all the electromagnetic radiation will move with the same velocity which is equal to the velocity of light.

The velocity of light c= 3×10^8 m/s.Hence the velocity of wave corresponding to this spectral line is 3×10^8 m/s.

We know that velocity of a wave is the product of frequency and wavelength of that corresponding wave. Mathematically we can write it as-

       v=\lambda f

[here \lambda is the wavelength of the wave ]

It is given that wavelength =21 cm=0.21 m

 The frequency is calculated as -

                                                         f=\frac{v}{\lambda}

                                                                 =\frac{c}{\lambda}

                                                                 =\frac{3*10^{8} m/s }{0.21 m}

                                                                  =14.2857*10^{8} s^{-1}

 Hence the antenna will be tuned to a frequency of 14.2857×10^8 Hz

Here Hertz[ Hz] is the unit of frequency.

7 0
3 years ago
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