Using the projectiles equations for y and x at what second does the ball have a vertical velocity of -39.2 m/s^2
1 answer:
Answer:
<h2>a) t = 4</h2>Explanation:
In projectile the time taken by an object to reach its maximum height is the maximum time (tmax).
tmax = Usin/g
U = velocity of the ball
= angle of projection = 90° (vertical velocity)
g = acceleration due to gravity = 9.8m/s²
tmax = Usin90°/9.8
tmax = U/9.8
T the maximum height U = Uy = 39.2 m/s²
tmax = 39.2/9.8
tmax = 4secs
To calculate the time at which the ball will have an horizontal velocity, we will use the same formula but will be 0° in this case
If Ux = Ucos
8 = Ucos0
U = 8m/s
Since t = Usin/g
t = 8sin0/9.8
t = 0sec
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Answer:
The index of refraction of the liquid is n = 1.33 equivalent to that of water
Explanation:
Solution:-
- The index of refraction of light in a medium ( n ) determines the degree of "bending" of light in that medium.
- The index of refraction is material property and proportional to density of the material.
- The denser the material the slower the light will move through associated with considerable diffraction angles.
- The lighter the material the faster the light pass through the material without being diffracted as much.
- So, in the other words index of refraction can be expressed as how fast or slow light passes through a medium.
- The reference of comparison of how fast or slow the light is the value of c = 3.0*10^8 m/s i.e speed of light in vacuum or also assumed to be the case for air.
- so we can mathematically express the index of refraction as a ratio of light speed in the material specified and speed of light.
- The light passes through a liquid with speed v = 2.25*10^8 m/s :
- The index of refraction of the liquid is n = 1.33 equivalent to that of water.