Answer:
1 / f = 1 / i + 1 / o thin lens equation
1 / i = 1 / f - 1 / o = (o - f) / (o * f)
i = o * f / (o - f)
i = 54.2 * 12.7 / (54.2 - 12.7) = 16.6 cm image distance
Image is real and inverted and 16.6 / 54.2 * 6 = 1.94 cm tall
A) 1973 I believe it had something to do with the Arab wars happening at the time
![v = 2.45×10^3\:\text{m/s}](https://tex.z-dn.net/?f=v%20%3D%202.45%C3%9710%5E3%5C%3A%5Ctext%7Bm%2Fs%7D)
Explanation:
Newton's 2nd Law can be expressed in terms of the object's momentum, in this case the expelled exhaust gases, as
(1)
Assuming that the velocity remains constant then
![F = \dfrac{d}{dt}(mv) = v\dfrac{dm}{dt}](https://tex.z-dn.net/?f=F%20%3D%20%5Cdfrac%7Bd%7D%7Bdt%7D%28mv%29%20%3D%20v%5Cdfrac%7Bdm%7D%7Bdt%7D)
Solving for
we get
![v = \dfrac{F}{\left(\frac{dm}{dt}\right)}\;\;\;\;\;\;\;(2)](https://tex.z-dn.net/?f=v%20%3D%20%5Cdfrac%7BF%7D%7B%5Cleft%28%5Cfrac%7Bdm%7D%7Bdt%7D%5Cright%29%7D%5C%3B%5C%3B%5C%3B%5C%3B%5C%3B%5C%3B%5C%3B%282%29)
Before we plug in the given values, we need to convert them first to their appropriate units:
The thrust <em>F</em><em> </em> is
![F = 7.5×10^6\:\text{lbs}×\dfrac{4.45\:\text{N}}{1\:\text{lb}} = 3.34×10^7\:\text{N}](https://tex.z-dn.net/?f=F%20%3D%207.5%C3%9710%5E6%5C%3A%5Ctext%7Blbs%7D%C3%97%5Cdfrac%7B4.45%5C%3A%5Ctext%7BN%7D%7D%7B1%5C%3A%5Ctext%7Blb%7D%7D%20%3D%203.34%C3%9710%5E7%5C%3A%5Ctext%7BN%7D)
The exhaust rate dm/dt is
![\dfrac{dm}{dt} = 15\dfrac{T}{s}×\dfrac{2000\:\text{lbs}}{1\:\text{T}}×\dfrac{1\:\text{kg}}{2.2\:\text{lbs}}](https://tex.z-dn.net/?f=%5Cdfrac%7Bdm%7D%7Bdt%7D%20%3D%2015%5Cdfrac%7BT%7D%7Bs%7D%C3%97%5Cdfrac%7B2000%5C%3A%5Ctext%7Blbs%7D%7D%7B1%5C%3A%5Ctext%7BT%7D%7D%C3%97%5Cdfrac%7B1%5C%3A%5Ctext%7Bkg%7D%7D%7B2.2%5C%3A%5Ctext%7Blbs%7D%7D)
![\;\;\;\;\;= 1.36×10^4\:\text{kg/s}](https://tex.z-dn.net/?f=%5C%3B%5C%3B%5C%3B%5C%3B%5C%3B%3D%201.36%C3%9710%5E4%5C%3A%5Ctext%7Bkg%2Fs%7D)
Therefore, the velocity at which the exhaust gases exit the engines is
![v = \dfrac{F}{\left(\frac{dm}{dt}\right)} = \dfrac{3.34×10^7\:\text{N}}{1.36×10^4\:\text{kg/s}}](https://tex.z-dn.net/?f=v%20%3D%20%5Cdfrac%7BF%7D%7B%5Cleft%28%5Cfrac%7Bdm%7D%7Bdt%7D%5Cright%29%7D%20%3D%20%5Cdfrac%7B3.34%C3%9710%5E7%5C%3A%5Ctext%7BN%7D%7D%7B1.36%C3%9710%5E4%5C%3A%5Ctext%7Bkg%2Fs%7D%7D)
![\;\;\;= 2.45×10^3\:\text{m/s}](https://tex.z-dn.net/?f=%5C%3B%5C%3B%5C%3B%3D%202.45%C3%9710%5E3%5C%3A%5Ctext%7Bm%2Fs%7D)
Answer:
During the voyage Charles Darwin explored the Galapagos islands and noticed the same species have different adaptations in places. ... Charles noticed that each species has the same ancestor but they evolve to adapt over time so they can live longer.
Explanation:
Try the solutions described in the attached picture, note the answers are marked with green colour.