1.add the amount of the diagram which is M+Y then dived the answer you get.
Answer:
Explanation:
The energy of a photon is given by the equation
, where h is the <em>Planck constant</em> and f the frequency of the photon. Thus, N photons of frequency f will give an energy of
.
We also know that frequency and wavelength are related by
, so we have
, where c is the <em>speed of light</em>.
We will want the number of photons, so we can write

We need to know then how much energy do we have to calculate N. The equation of power is
, so for the power we have and considering 1 second we can calculate the total energy, and then only consider the 4% of it which will produce light, or better said, the N photons, which means it will be
.
Putting this paragraph in equations:
.
And then we can substitute everything in our equation for number of photons, in S.I. and getting the values of constants from tables:

Answer:
a) 17.33 V/m
b) 6308 m/s
Explanation:
We start by using equation of motion
s = ut + 1/2at², where
s = 1.2 cm = 0.012 m
u = 0 m/s
t = 3.8*10^-6 s, so that
0.012 = 0 * 3.8*10^-6 + 0.5 * a * (3.8*10^-6)²
0.012 = 0.5 * a * 1.444*10^-11
a = 0.012 / 7.22*10^-12
a = 1.66*10^9 m/s²
If we assume the electric field to be E, and we know that F =qE. Also, from Newton's law, we have F = ma. So that, ma = qE, and E = ma/q, where
E = electric field
m = mass of proton
a = acceleration
q = charge of proton
E = (1.67*10^-27 * 1.66*10^9) / 1.6*10^-19
E = 2.77*10^-18 / 1.6*10^-19
E = 17.33 V/m
Final speed of the proton can be gotten by using
v = u + at
v = 0 + 1.66*10^9 * 3.8*10^-6
v = 6308 m/s
Answer:
The range of powers is 
Explanation:
From the question we are told that
The far point of the left eye is 
The near point of the left eye is 
The near point with the glasses on is 
From these parameter we can see that with the glass on that for near point the
Object distance would be 
Image distance would be 
To obtain the focal length we would apply the lens formula which is mathematically represented as

substituting values


converting to meters


Generally the power of the lens is mathematically represented as

Substituting values


From these parameter we can see that with the glass on that for far point the
Object distance would be 
Image distance would be 
To obtain the focal length of the lens we would apply the lens formula which is mathematically represented as

substituting values


converting to meters

Generally the power of the lens is mathematically represented as

Substituting values


This implies that the range of powers of the lens in his glass is
