wave function of a particle with mass m is given by ψ(x)={ Acosαx −
π
2α
≤x≤+
π
2α
0 otherwise , where α=1.00×1010/m.
(a) Find the normalization constant.
(b) Find the probability that the particle can be found on the interval 0≤x≤0.5×10−10m.
(c) Find the particle’s average position.
(d) Find its average momentum.
(e) Find its average kinetic energy −0.5×10−10m≤x≤+0.5×10−10m.
Answer:
Explanation:
For this problem we use the translational equilibrium condition. Our reference frame for block 1 is one axis parallel to the plane and the other perpendicular to the plane.
X axis
-Aₓ - f_e +T = 0 (1)
Y axis
N₁ - W_y = 0 ( 2)
let's use trigonometry for the weight components
sin θ = Wₓ / W
cos θ = W_y / W
Wₓ = W sin θ
W_y = W cos θ
We write the diagram for the second body.
Note that in the block the positive direction rd upwards, therefore for block 2 the positive direction must be downwards
W₂ -T = 0 (3)
we add the equations is 1 and 3
- W₁ sin θ - μ N₁ + W₂ = 0
from equation 2
N₁ = W₁ cos θ
we substitute
-W₁ sin θ - μ (W₁ cos θ) + W₂ = 0
W₂ = m₁ g (without ea - very expensive)
This is the smallest value that supports the equilibrium system
Answer:
The resistivity of the material used to make the rod is ρ= 7.5 * 10⁻⁷ Ω.m
Explanation:
R= 0.2 Ω
L= 0.8 m
S= 1.5mm*2mm= 3 mm² = 3 * 10⁻⁶ m²
ρ = (R*S)/L
ρ= 7.5 * 10⁻⁷ Ω.m
Answer:
29.4 N/m
0.1
Explanation:
a) From the restoring Force we know that :
F_r = —k*x
the gravitational force :
F_g=mg
Where:
F_r is the restoring force .
F_g is the gravitational force
g is the acceleration of gravity
k is the constant force
xi , x2 are the displacement made by the two masses.
Givens:
<em>m1 = 1.29 kg</em>
<em>m2 = 0.3 kg </em>
<em>x1 = -0.75 m </em>
<em>x2 = -0.2 m </em>
<em>g = 9.8 m/s^2 </em>
Plugging known information to get :
F_r =F_g
-k*x1 + k*x2=m1*g-m2*g
k=29.4 N/m
b) To get the unloaded length 1:
l=x1-(F_1/k)
Givens:
m1 = 1.95kg , x1 = —0.75m
Plugging known infromation to get :
l= x1 — (F_1/k)
= 0.1
