Answer:
r = 71.8⁰
Explanation:
given,
refractive index of the glass 1 = 1.70
refractive index of glass 2 = 1.58
angle of incidence = 62°
angle of refraction =?
using Snell's law
![\dfrac{sin\ i}{sin\ r} = \dfrac{n_2}{n_1}](https://tex.z-dn.net/?f=%5Cdfrac%7Bsin%5C%20i%7D%7Bsin%5C%20r%7D%20%3D%20%5Cdfrac%7Bn_2%7D%7Bn_1%7D)
![\dfrac{sin\ 62^0}{sin\ r} = \dfrac{1.58}{1.70}](https://tex.z-dn.net/?f=%5Cdfrac%7Bsin%5C%2062%5E0%7D%7Bsin%5C%20r%7D%20%3D%20%5Cdfrac%7B1.58%7D%7B1.70%7D)
1.7 ×sin 62 ^0 = 1.58× sin r
![sinr = \dfrac{1.7\times sin 62^0}{1.58}](https://tex.z-dn.net/?f=sinr%20%3D%20%5Cdfrac%7B1.7%5Ctimes%20sin%2062%5E0%7D%7B1.58%7D)
sin r = 0.95
r = sin⁻¹(0.95)
r = 71.8⁰
angle of refraction =r = 71.8⁰
I have the same question I can't get ut
<u>Momentum</u>
- a vector quantity; has both magnitude and direction
- has the same direction as object's velocity
- can be represented by components x & y.
Find linebacker momentum given m₁ = 120kg, v₁ = 8.6 m/s north
P₁ = m₁v₁
P₁ = (120)(8.6)
[ P₁ = 1032 kg·m/s ] = y-component, linebacker momentum
Find halfback momentum given m₂ = 75kg, v₂ = 7.4 m/s east
P₂ = m₂v₂
P₂ = (75)(7.4)
[ P₂ = 555 kg·m/s ] = x-component, halfback momentum
Find total momentum using x and y components.
P = √(P₁)² + (P₂)²
P = √(1032)² + (555)²
[[ P = 1171.77 kg·m/s ]] = magnitude
!! Finally, to find the magnitude of velocity, take the divide magnitude of momentum by the total mass of the players.
P = mv
P = (m₁ + m₂)v
1171.77 = (120 + 75)v <em>[solve for v]</em>
<em />v = 1171.77/195
v = 6.0091 ≈ 6.0 m/s
If asked to find direction, take inverse tan of x and y components.
tanθ = (y/x)
θ = tan⁻¹(1032/555)
[ θ = 61.73° north of east. ]
The magnitude of the velocity at which the two players move together immediately after the collision is approximately 6.0 m/s.
<span>A center-seeking force related to acceleration is centripetal force. The answer is letter A. The rest of the choices do not answer the question above.</span>
Answer:
A
Explanation:
heat energy can be converted into electricity with very high efficiency. .