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Oliga [24]
3 years ago
7

Two arrows are shot vertically upward. the second arrow is shot after the first one, but while the first is still on its way up.

the initial speeds are such that both arrows reach their maximum heights at the same instant, although these heights are different. suppose that the initial speed of the first arrow is 31.8 m/s and that the second arrow is fired 1.82 s after the first. determine the initial speed of the second arrow.
Physics
1 answer:
Neko [114]3 years ago
7 0
<span>Now that you know the time to reach its maximum height, you have enough information to find out the initial velocity of the second arrow. Here's what you know about it: its final velocity is 0 m/s (at the maximum height), its time to reach that is 2.8 seconds, but wait! it was fired 1.05 seconds later, so take off 1.05 seconds so that its time is 1.75 seconds, and of course gravity is still the same at -9.8 m/s^2. Plug those numbers into the kinematic equation (Vf=Vi+a*t, remember?) for 0=Vi+-9.8*1.75 and solve for Vi to get....... 17.15 m/s</span>
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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.

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