Answer:
a) P=0.25x10^-7
b) R=B*N2*E
c) N=1.33x10^9 photons
Explanation:
a) the spontaneous emission rate is equal to:
1/tsp=1/3 ms
the stimulated emission rate is equal to:
pst=(N*C*o(v))/V
where
o(v)=((λ^2*A)/(8*π*u^2))g(v)
g(v)=2/(π*deltav)
o(v)=(λ^2)/(4*π*tp*deltav)
Replacing values:
o(v)=0.7^2/(4*π*3*50)=8.3x10^-19 cm^2
the probability is equal to:
P=(1000*3x10^10*8.3x10^-19)/(100)=0.25x10^-7
b) the rate of decay is equal to:
R=B*N2*E, where B is the Einstein´s coefficient and E is the energy system
c) the number of photons is equal to:
N=(1/tsp)*(V/C*o)
Replacing:
N=100/(3*3x10^10*8.3x10^-19)
N=1.33x10^9 photons
Answer: C
Combustion
Explanation:
Oxygen support combustion in a chemical reaction. Oxygen support burning.
In the absence of oxygen, combustion reaction can never take place. And when the supply of oxygen is being cut off, combustion reaction will automatically stop.
The correct option is C which is combustion
In order for a system to be in equilibrium , two conditions must be met. Net force must be 0.
151.9j
Explanation:
PE=1/2kx^2
PE=1/2(980)(.50)= 245j
PE=(1/2)(980)(.81)= 396.9j
396.9- 245= 151.9j
Answer:
4.53 second, 195 ft/s
Explanation:
u = 50 ft/s
h = 555 ft
g = 32 ft/s^2
Let the time taken by the ball to reach the ground is t and the velocity of the ball as it hits the ground is v.
Use third equation of motion
v = 195 ft /s
Thus, the ball strikes the ground with velocity 195 ft/s.
Use first equation of motion
v = u + at
195 = 50 + 32 t
32 t = 145
t = 4.53 second
Thus, the ball reach the ground in 4.53 second.