Answer:
3 photons
Explanation:
The energy of a photon E can be calculated using this formula:
Where 
corresponds to Plank constant (6.626070x10^-34Js), 
is the speed of light in the vacuum (299792458m/s) and 
is the wavelength of the photon(in this case 800nm).
Tranform the units
The band Gap is 4eV, divide the band gap between the energy of the photon:
Rounding to the next integrer: 3.
Three photons are the minimum to equal or exceed the band gap.
Answer:
Well the definition of an application is the act of putting to a special use or purpose so lam assuming that you want specific uses that scientists make of gravity in their work.
Well our first application has helped us to send satellites around the solar system with what Nasa calls gravity assist. Using a particular planets gravity to slingshot a satellite to another destination. Look it up.
The next application much simpler but here on Earth. There are many hydro-electric power stations in use all over the world. Water is stored at a high level and released falling 100s of metres to a turbine where it generates electricity.
Hope that helps.
Explanation:
Answer:
Explanation:
a. The equation of Lorentz transformations is given by:
x = γ(x' + ut')
x' and t' are the position and time in the moving system of reference, and u is the speed of the space ship. x is related to the observer reference.
x' = 0
t' = 5.00 s
u =0.800 c,
c is the speed of light = 3×10⁸ m/s
Then,
γ = 1 / √ (1 - (u/c)²)
γ = 1 / √ (1 - (0.8c/c)²)
γ = 1 / √ (1 - (0.8)²)
γ = 1 / √ (1 - 0.64)
γ = 1 / √0.36
γ = 1 / 0.6
γ = 1.67
Therefore, x = γ(x' + ut')
x = 1.67(0 + 0.8c×5)
x = 1.67 × (0+4c)
x = 1.67 × 4c
x = 1.67 × 4 × 3×10⁸
x = 2.004 × 10^9 m
x ≈ 2 × 10^9 m
Now, to find t we apply the same analysis:
but as x'=0 we just have:
t = γ(t' + ux'/c²)
t = γ•t'
t = 1.67 × 5
t = 8.35 seconds
b. Mavis reads 5 s on her watch which is the proper time.
Stanley measured the events at a time interval longer than ∆to by γ,
such that
∆t = γ ∆to = (5/3)(5) = 25/3 = 8.3 sec which is the same as part (b)
c. According to Stanley,
dist = u ∆t = 0.8c (8.3) = 2 x 10^9 m
which is the same as in part (a)
Protons are positive,
electrons are negative,
and neutrons are neutral.
In the nucleus, there are protons and neutrons, so the charge of a nucleus is positive.
