Answer:
#_photons = 30 photons / s
Explanation:
Let's start by finding the energy of a photon of light, let's use the Planck relation
E = h f
the speed of light is related to wavelength and frequency
c = λ f
we substitute
E = h c /λ
E₀ = 6.63 10⁻³⁴ 3 10⁸/500 10⁻⁹
E₀ = 3.978 10⁻¹⁹ J
now let's use a direct proportion rule. If the energy of a photon is Eo, how many fornes has an energy E = 1.2 10⁻¹⁷ J in a second
#_photons = 1 photon (E / Eo)
#_photons = 1 1.2 10⁻¹⁷ /3.978 10⁻¹⁹
#_photons = 3.0 10¹
#_photons = 30 photons / s
Explanation:
Given that,
Mass if the rock, m = 1 kg
It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.
We need to find the mass of the meter stick. The force acting by the stone is
F = 1 × 9.8 = 9.8 N
Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.

W = 3.266 N
The mass of the meters stick is :

So, the mass of the meter stick is 0.333 kg.
Answer:
We would not be able to make our way around the earth's surface, to read, to sculpt or to create technology because we cannot see!
Explanation:
The minimum angular separation that can be distinguished by an eye gives the angular resolution of the eye.
Given that the Angular resolution with infrared radiation is =
equal to 
This resolution is very much greater than that of the eye 
The angular resolution that our eyes can see is about
at arms length
Angular resolution of infrared =
at arms length
We therefore cannot read, sculpt or create technology because we cannot see.
By the use of "Solar cells" or "Solar panel", we can trap solar energy in silicon/germanium rods, then we can convert in into electricity too.
Hope this helps!
Given that,
At 3.00 m from a source that is emitting sound uniformly in all directions, the sound level is 60.0 dB.
To find,
The distance from the source would the sound level be one-fourth the sound level at 3.00 m.
Solution,
The intensity from a source is inversely proportional to the distance.
Let I₁ = 60 dB, r₁ = 3 m, I₂ = 60/4 = 15 dB, r₂ =?
Using relation :

So, at a distance of 6 m the sound level will be one fourth of the sound level at 3 m.