Answer:
n_cladding = 1.4764
Explanation:
We are told that θ_max = 5 °
Thus;
θ_max + θ_c = 90°
θ_c = 90° - θ_max
θ_c = 90° - 5°
θ_c = 85°
Now, critical angle is given by;
θ_c = sin^(-1) (n_cladding/n_core)
sin θ_c = (n_cladding/n_core)
n_cladding = (n_core) × sin θ_c
Plugging in the relevant values, we have;
n_cladding = 1.482 × sin 85
n_cladding = 1.4764
Answer:
Work done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is
Explanation:
The potential energy is given by:
U=Q*V
where:
Q is the charge
V is the potential difference
Potential Difference=V=
So,
Where:
k is Coulomb Constant=
q is the charge on electron=
r is the distance=
For 3 Electrons Potential Energy or work Done is:
Work done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is
Answer:
Give the child a lot of room to your side, which may mean moving closer to the oncoming vehicles.
Explanation:
I majored in Physics.
Answer:
The energy dissipated as the puck slides over the rough patch is 1.355 J
Explanation:
Given;
mass of the hockey puck, m = 0.159 kg
initial speed of the puck, u = 4.75 m/s
final speed of the puck, v = 2.35 m/s
The energy dissipated as the puck slides over the rough patch is given by;
ΔE = ¹/₂m(v² - u²)
ΔE = ¹/₂ x 0.159 (2.35² - 4.75²)
ΔE = -1.355 J
the lost energy is 1.355 J
Therefore, the energy dissipated as the puck slides over the rough patch is 1.355 J
Answer:
v_avg = 37 km/h
Explanation:
To find the average velocity in the complete trajectory you use the following formula:
( 1 )
v1: velocity in the first part of the trajectory = 70 km/h
v2: velocity in the second part of the trajectory = ?
You calculate v2 by using the following equation for a motion with constant velocity:
you replace the values of v1 and v2 in (1) and you obtain:
hence, the average velocity is 37 km/h