Answer:
![\frac{M_e}{M_s} = 3.07 \times 10^{-6}](https://tex.z-dn.net/?f=%5Cfrac%7BM_e%7D%7BM_s%7D%20%3D%203.07%20%5Ctimes%2010%5E%7B-6%7D)
Explanation:
As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula
![T = 2\pi \sqrt{\frac{r^3}{GM}}](https://tex.z-dn.net/?f=T%20%3D%202%5Cpi%20%5Csqrt%7B%5Cfrac%7Br%5E3%7D%7BGM%7D%7D)
now for the time period of moon around the earth we can say
![T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}](https://tex.z-dn.net/?f=T_1%20%3D%202%5Cpi%5Csqrt%7B%5Cfrac%7Br_1%5E3%7D%7BGM_e%7D%7D)
here we know that
![T_1 = 0.08 year](https://tex.z-dn.net/?f=T_1%20%3D%200.08%20year)
![r_1 = 0.0027 AU](https://tex.z-dn.net/?f=r_1%20%3D%200.0027%20AU)
= mass of earth
Now if the same formula is used for revolution of Earth around the sun
![T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}](https://tex.z-dn.net/?f=T_2%20%3D%202%5Cpi%5Csqrt%7B%5Cfrac%7Br_2%5E3%7D%7BGM_s%7D%7D)
here we know that
![r_2 = 1 AU](https://tex.z-dn.net/?f=r_2%20%3D%201%20AU)
![T_2 = 1 year](https://tex.z-dn.net/?f=T_2%20%3D%201%20year)
= mass of Sun
now we have
![\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}](https://tex.z-dn.net/?f=%5Cfrac%7BT_2%7D%7BT_1%7D%20%3D%20%5Csqrt%7B%5Cfrac%7Br_2%5E3%20M_e%7D%7Br_1%5E3%20M_s%7D%7D)
![\frac{1}{0.08} = \sqrt{\frac{1 M_e}{(0.0027)^3M_s}}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B0.08%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B1%20M_e%7D%7B%280.0027%29%5E3M_s%7D%7D)
![12.5 = \sqrt{(5.08 \times 10^7)\frac{M_e}{M_s}}](https://tex.z-dn.net/?f=12.5%20%3D%20%5Csqrt%7B%285.08%20%5Ctimes%2010%5E7%29%5Cfrac%7BM_e%7D%7BM_s%7D%7D)
![\frac{M_e}{M_s} = 3.07 \times 10^{-6}](https://tex.z-dn.net/?f=%5Cfrac%7BM_e%7D%7BM_s%7D%20%3D%203.07%20%5Ctimes%2010%5E%7B-6%7D)
The speed at which sound travels through the gas in the tube is 719.94m/s
<u>Explanation:</u>
Given:
Frequency, f = 11999Hz
Wavelength, λ = 0.03m
Velocity, v = ?
Sound speed in the tube is calculated by multiplying the frequency v by the wavelength λ.
As the sound loudness changed from a maximum to a minimum, then we know the sound interference in the case changed from constructive interference (the two sound waves are in phase, i.e. peaks are in a line with peaks and so the troughs), to a destructive interference (peaks coinciding with troughs). The least distance change required to cause such a change is a half wavelength distance, so:
λ/2 = 0.03/2
λ = 0.06m
We know,
v = λf
v = 0.06 X 11999Hz
v = 719.94m/s
Therefore, the speed at which sound travels through the gas in the tube is 719.94m/s
From the information given,
diameter of ornament = 8
radius = diameter/2 = 8/2
radius of curvature, r = 4
Recall,
focal length, f = radius of curvature/2 = 4/2
f = 2
Recall,
magnification = image d