Answer:
3 photons
Explanation:
The energy of a photon E can be calculated using this formula:
![E=\frac{hc}{\lambda}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7Bhc%7D%7B%5Clambda%7D)
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).
![E=\frac{hc}{\lambda}=\frac{(6.626070\times10^{-34})(299792458)}{800\times10^{-9}}=\frac{1.986445812\times10^-25}{800}=2.483057265\times10^{-19}J](https://tex.z-dn.net/?f=E%3D%5Cfrac%7Bhc%7D%7B%5Clambda%7D%3D%5Cfrac%7B%286.626070%5Ctimes10%5E%7B-34%7D%29%28299792458%29%7D%7B800%5Ctimes10%5E%7B-9%7D%7D%3D%5Cfrac%7B1.986445812%5Ctimes10%5E-25%7D%7B800%7D%3D2.483057265%5Ctimes10%5E%7B-19%7DJ)
Tranform the units
![1eV=1.602176634\times10^{-19}J\\2.483057265\times10^{-19}J(\frac{1eV}{1.602176634\times10^{-19}J})=1.549802445eV](https://tex.z-dn.net/?f=1eV%3D1.602176634%5Ctimes10%5E%7B-19%7DJ%5C%5C2.483057265%5Ctimes10%5E%7B-19%7DJ%28%5Cfrac%7B1eV%7D%7B1.602176634%5Ctimes10%5E%7B-19%7DJ%7D%29%3D1.549802445eV)
The band Gap is 4eV, divide the band gap between the energy of the photon:
![\frac{4ev}{1.549802445eV}=2.508974118](https://tex.z-dn.net/?f=%5Cfrac%7B4ev%7D%7B1.549802445eV%7D%3D2.508974118)
Rounding to the next integrer: 3.
Three photons are the minimum to equal or exceed the band gap.
Answer:
96w
Explanation:
p=Iv..where v=12 and I=8.0
The answer is parallel
If the <span>circuits in a car</span> were series, they would go out at the same time.
I hope this helps! :3
Let's break the question into two parts:
1) The force needed in Ramp scenario.
2) The effort force needed in the lever scenario.
1. Ramp Scenario: In an incline, the only component of cart's weight(
mg) that is in the direction of motion is
. Therefore the effort force in this case must be equal or greater than
.
Now we need to find
![\alpha](https://tex.z-dn.net/?f=%20%5Calpha%20%20)
.
![\alpha](https://tex.z-dn.net/?f=%20%5Calpha%20)
is the angle between the incline of the ramp and the ground.
Since the height is
5m and the length of the ramp is
8m, ![sin \alpha](https://tex.z-dn.net/?f=sin%20%5Calpha%20%20)
would be
5/8 or 0.625. Now that you have
![sin \alpha](https://tex.z-dn.net/?f=sin%20%5Calpha%20)
, mutiple it with
mg.
=> m*g*
![sin \alpha](https://tex.z-dn.net/?f=sin%20%5Calpha%20)
= 20 * 10 * 5 / 8. (Taking g = 10 m/s² for simplicity) = 125N
Therefore, the minimum Effort force you would require in this case is
125N.
2. Lever Scenario:
Just apply "moment action" in this case, which is:
![F_{e} d_{e} = F_{r} d_{r}](https://tex.z-dn.net/?f=F_%7Be%7D%20%20d_%7Be%7D%20%20%3D%20F_%7Br%7D%20%20d_%7Br%7D)
![F_{e}](https://tex.z-dn.net/?f=F_%7Be%7D%20)
= ?
![F_{r}](https://tex.z-dn.net/?f=F_%7Br%7D%20)
= mg = 20 * 10 = 200N
![d_{e}](https://tex.z-dn.net/?f=d_%7Be%7D%20)
= 10m
![d_{r}](https://tex.z-dn.net/?f=d_%7Br%7D%20)
= 1m
Plug-in the values in the above equation:
![F_{e}](https://tex.z-dn.net/?f=F_%7Be%7D%20)
= 200/10=
20NAs 20N << 125N, the best choice is to use lever.