Answer: 3217.79 hours.
Explanation:
Given, A 140 lb. climber saved her potential energy as she descended from Mt. Everest (Elev. 29,029 ft) to Kathmandu (Elev. 4,600 ft).
Power = 0.4 watt
Mass of climber = 140 lb
= 140 x 0.4535 kg [∵ 1 lb= 0.4535 kg]
⇒ Mass of climber (m) = 63.50 kg
Let
and ![h_2= 4,600 ft = 1402.08\ m](https://tex.z-dn.net/?f=h_2%3D%204%2C600%20ft%20%3D%201402.08%5C%20m)
Now, Energy saved =![mg(h_1-h_2)=(63.50)(9.8)(8848.04-1402.08)=4633620.91\ J](https://tex.z-dn.net/?f=mg%28h_1-h_2%29%3D%2863.50%29%289.8%29%288848.04-1402.08%29%3D4633620.91%5C%20J)
![\text{Power}=\dfrac{\text{energy}}{\text{time}}\\\\\Rightarrow 0.4=\dfrac{4633620.91}{\text{time}}\\\\\Rightarrow\ \text{time}=\dfrac{4633620.91}{0.4}\approx11584052.28\text{ seconds}\\\\=\dfrac{11584052.28}{3600}\text{ hours}\ \ \ [\text{1 hour = 3600 seconds}]\\\\=3217.79\text{ hours}](https://tex.z-dn.net/?f=%5Ctext%7BPower%7D%3D%5Cdfrac%7B%5Ctext%7Benergy%7D%7D%7B%5Ctext%7Btime%7D%7D%5C%5C%5C%5C%5CRightarrow%200.4%3D%5Cdfrac%7B4633620.91%7D%7B%5Ctext%7Btime%7D%7D%5C%5C%5C%5C%5CRightarrow%5C%20%5Ctext%7Btime%7D%3D%5Cdfrac%7B4633620.91%7D%7B0.4%7D%5Capprox11584052.28%5Ctext%7B%20seconds%7D%5C%5C%5C%5C%3D%5Cdfrac%7B11584052.28%7D%7B3600%7D%5Ctext%7B%20hours%7D%5C%20%5C%20%5C%20%5B%5Ctext%7B1%20hour%20%3D%203600%20seconds%7D%5D%5C%5C%5C%5C%3D3217.79%5Ctext%7B%20hours%7D)
Hence, she can power her 0.4 watt flashlight for 3217.79 hours.
It's located in the nucleus
Buhrs atomic model differed from ruthofords because it explained that electrons exist in specified energy levels surrounding the nucleus. This means that, Ruthoford believed that electrons can't do very much. However, Buhrs' model showed that electrons are much more powerful than anyone else believes they can be.
Explanation:
Given that,
Distance 1, r = 100 m
Intensity, ![I_1=1.24\times 10^{-8}\ W/m^2](https://tex.z-dn.net/?f=I_1%3D1.24%5Ctimes%2010%5E%7B-8%7D%5C%20W%2Fm%5E2)
If distance 2, r' = 25 m
We need to find the intensity and the intensity level at 25 meters. Intensity and a distance r is given by :
.........(1)
Let I' is the intensity at r'. So,
............(2)
From equation (1) and (2) :
![I'=\dfrac{Ir}{r'^2}](https://tex.z-dn.net/?f=I%27%3D%5Cdfrac%7BIr%7D%7Br%27%5E2%7D)
![I'=\dfrac{1.24\times 10^{-8}\times 100}{25^2}](https://tex.z-dn.net/?f=I%27%3D%5Cdfrac%7B1.24%5Ctimes%2010%5E%7B-8%7D%5Ctimes%20100%7D%7B25%5E2%7D)
![I'=1.98\times 10^{-9}\ W/m^2](https://tex.z-dn.net/?f=I%27%3D1.98%5Ctimes%2010%5E%7B-9%7D%5C%20W%2Fm%5E2)
Intensity level is given by :
, ![I_o=10^{-12}\ W/m^2](https://tex.z-dn.net/?f=I_o%3D10%5E%7B-12%7D%5C%20W%2Fm%5E2)
![dB=10\ log(\dfrac{1.98\times 10^{-9}}{10^{-12}})](https://tex.z-dn.net/?f=dB%3D10%5C%20log%28%5Cdfrac%7B1.98%5Ctimes%2010%5E%7B-9%7D%7D%7B10%5E%7B-12%7D%7D%29)
dB = 32.96 dB
Hence, this is the required solution.
Well according to the molecular formula of glucose, one molecule would have 6 carbon atoms, and thus 2 molecules of glucose would have 12 carbon atoms.
The correct response would be B. 12.