it is the point at infinity where it is at a distance from the curve equal to the radius of curvature lying on the normal vector. Sorry no diagram
Answer: Their u go i found it their was about 3 pages i did not no what pages u had to do.
Explanation:
Answer:
Light's angle of refraction = 37.1° (Approx.)
Explanation:
Given:
Index of refraction = 1.02
Base of refraction = 1
Angle of incidence = 38°
Find:
Light's angle of refraction
Computation:
Using Snell's law;
Sin[Angle of incidence] / Sin[Light's angle of refraction] = Index of refraction / Base of refraction
Sin38 / Light's angle of refraction = 1.02 / 1
Sin[Light's angle of refraction] = Sin 38 / 1.02
Sin[Light's angle of refraction] = [0.6156] / 1.02
Sin[Light's angle of refraction] = 0.6035
Light's angle of refraction = 37.1° (Approx.)
Answer:
2.2 m/s^2
Explanation:
Acceleration = Force / Mass
= 7.92 / 3.6 = 2.2m/s^2
Hope this help you :3
The answer is d I took the test