Answer:
The answer for a classical particle is 0.00595
Explanation:
The equation of the wave function of a particle in a box in the second excited state equals:
ψ(x) = ((2/L)^1/2) * sin((3*pi*x)/L)
The probability is equal to:
P(x)dx = (|ψ(x)|^2)dx = ((2/L)^1/2) * sin((3*pi*x)/L) = (2/L) * sin^2((3*pi*x)/L) dx
for x = 0.166 nm
P(x)dx = (2/0.167) * sin^2((3*pi*0.166)/0.167) * 100 pm = 0.037x10^-3
for x = 0.028 nm
P(x)dx = (2/0.167) * sin^2((3*pi*0.028)/0.167) * 100 pm = 11x10^-3
for x = 0.067 nm
P(x)dx = (2/0.167) * sin^2((3*pi*0.067)/0.167) * 100 pm = 3.99x10^-3
therefore, the classical probability is equal to:
(1/L)dx = (1/0.167)*100 pm = 0.00595
I had this question before can you show me the answers choices
Answer:
3.2 m/s
Explanation:
Given:
Δx = 1000 m
v₀ = 23 m/s
a = -0.26 m/s²
t = 76 s
Find: v
This problem is over-defined. We only need 3 pieces of information, and we're given 4. There are several equations we can use. For example:
v = at + v₀
v = (-0.26 m/s²) (76 s) + (23 m/s)
v = 3.2 m/s
Or:
Δx = ½ (v + v₀) t
(1000 m) = ½ (v + 23 m/s) (76 s)
v = 3.3 m/s
Or:
v² = v₀² + 2aΔx
v² = (23 m/s)² + 2(-0.26 m/s²)(1000 m)
v = 3.0 m/s
Or:
Δx = vt − ½ at²
(1000 m) = v (76 s) − ½ (-0.26 m/s²) (76 s)²
v = 3.3 m/s
As you can see, you get slightly different answers depending on which variables you use. Since 1000 m has 1 significant figure, compared to the other variables which have 2 significant figures, I recommend using the first equation.
Answer:
the lowest velocity which a body must have in order to escape the gravitational attraction of a particular planet or other object.
Explanation: