-- The energy of one photon is <em>(h · frequency of the light)</em>
' h ' is 6.626 × 10⁻³⁴ m²-kg/s ("Planck's Constant")
-- The question doesn't tell you the frequency of the light from the LED, but it tells you the wavelength, and
<em>Frequency = (speed of light) / (wavelength) </em>.
-- Now you have everything you need to calculate the <em>energy carried by one photon from the LED</em>.
-- The power of the light from the LED is 120 milliwatts. That's <em>0.120 Joule of energy per second</em>.
Now you should be able to find the number of photons per second. It's going to be <em>(0.120 Joule) / (energy carried by one photon)</em> .
When I scribbled it out on a scrap of scratch paper, I got 3.853 x 10³⁸ photons, but you'd better really check that out.
Answer:


Explanation:
<u>Given:</u>
No. of times = 600 times
Time = 60 seconds
<u>Required:</u>
Frequency = f = ?
Time Period = T = ?
<u>Solution:</u>
<h2>Frequency = No. of Times / Time</h2>
Frequency = 600 / 60
<u>Frequency = 10 Hz</u>
<h2>Time Period = 1 / Frequency</h2>
Time Period = 1 / 10
<u>Time Period = 0.1 secs</u>
Picture? I may be able to answer if you have a chart or some kind of graph as a referral to the question
Answer:
The soil plays an important part in the storage and distribution of the water that reaches it. The role the soil plays in the hydrological cycle is crucial for the maintenance of the biosphere, because it is provides water for plants, animals and humans.
Answer:
a) velocity v = 322.5m/s
b) time t = 19.27s
Explanation:
Note that;
ads = vdv
where
a is acceleration
s is distance
v is velocity
Given;
a = 6 + 0.02s
so,

Remember that
![v = \frac{ds}{dt} \\\frac{ds}{v} = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^{2} } } } \, ds = \int\limits^t_0 {} \, dt \\t= (5\sqrt{2} ) ln \frac{| [s + 300 + \sqrt{(s^{2} + 600s)} ] |}{300} .......2](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7Bds%7D%7Bdt%7D%20%5C%5C%5Cfrac%7Bds%7D%7Bv%7D%20%3D%20dt%5C%5C%5Cint%5Climits%5Es_0%20%7B%5Cfrac%7Bds%7D%7B%5Csqrt%7B12s%2B0.02s%5E%7B2%7D%20%7D%20%7D%20%7D%20%5C%2C%20ds%20%3D%20%5Cint%5Climits%5Et_0%20%7B%7D%20%5C%2C%20dt%20%5C%5Ct%3D%20%20%285%5Csqrt%7B2%7D%20%29%20ln%20%20%5Cfrac%7B%7C%20%5Bs%20%2B%20300%20%2B%20%5Csqrt%7B%28s%5E%7B2%7D%20%20%2B%20600s%29%7D%20%5D%20%7C%7D%7B300%7D%20.......2)
substituting s = 2km =2000m, into equation 1
v = 322.5m/s
substituting s = 2000m into equation 2
t = 19.27s