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<h2>Answer:The angle between the refracted ray and the normal.</h2>
Explanation:
Refraction is defined as the phenomenon of bending or changing course of a light ray when it passes from one medium to another medium.
The new ray formed after the bending at the surface of new medium is called the refracted ray.
The angle between the refracted ray and the normal of the point on the surface of the new medium where the ray enters is called angle of refraction.
Refer the attachment for instance.
In short,it is the angle between the refracted ray and the normal.
Answer:
a) See attachment.
b) a_avg = -8.66 i - 2.33 j
c) a = 195 degrees from + x-axis. |a_avg| = 8.97 m/s^2
Explanation:
Given:
- Initial Velocity V_i = (90 i + 110 j)
- Final velocity V_f = (-170 i + 40 j)
- t_i = 0
- t_f = 30.0 s
Find:
(a) Sketch the velocity vectors at t1 and t2. How do these two vectors differ?
(b) the components of the average acceleration
(c) the magnitude and direction of the average acceleration.
Solution:
- The change in velocity is given by:
Δ V = (V_f - V_i)
Δ V = ( -170 i - 90 i + 40 j - 110 j)
Δ V = (-260 i - 70 j)
- The change in time is given as:
Δ t = t_f - t_i = 30 - 0 = 30 s
-The average acceleration is:
a_avg = Δ V / Δ t
a_avg = (-260 i - 70 j) / 30
a_avg = -8.66 i - 2.33 j
- The magnitude of acceleration is:
|a_avg| = sqrt ( 8.66^2 + 2.33^2)
|a_avg| = sqrt (80.4245)
|a_avg| = 8.97 m/s^2
- The direction of acceleration is given by:
tan (Q) = a_y / a_x
Q = arctan(a_y / a_x)
Q = arctan(2.33 / 8.66)
Q = 15 degrees
Hence, 180 + 15 = a. a = 195 degrees from + x-axis.
Answer:
The ball never passes 2m high, Hmax=1.27m
Explanation:
we assume the ball doesn't bounce when it hits the ground.
We calculate the maximum height, Vf = 0.
So, the ball never passes 2m high.
Kinematics equations:
Find annexed the graphics of x(t) and v(t)
