Light is shining perpendicularly on the surface of the earth with an intensity of 680 w/m^2. assuming that all the photons in the light have the same wavelength (in vacuum) of 678 nm, determine the number of photons per second per square meter that reach the earth.

3.066×10^21  photons/(s.m^2)

Explanation:

The power per area is:

Power/A = (# of photons /t /A)×(energy / photon)

E/photons = h×c/(λ)

photons /t /A = (Power/A)×λ /(h×c)  

photons /t /A = (P/A)×λ/(hc)

photons /t /A = (680)×(678×10^-9)/(6.63×10^-34)×(3×10^-8)

                      = 3.066×10^21

Therefore, the number of photons per second per square meter 3.066×10^21  photons/(s.m^2).

uranium fissions/breaks up giving out energy and neutrons

a=d/t

a="triangle"v/t

explanation:

a = acceleration