Answer: the angle of refraction in the water is 34.2 degrees and then in the glass is 41 degrees approximately
Explanation:
A laser light travels from air (n=1) through water (n=1.33), and then a piece of glass (n=1.55).
If the light first enters the water at an angle of 25.0 degrees (from air to water). That means
The incident angle = 25 degrees
Using Snell' law
n(air)/n(water) = sin 25/sin R
Where R is the angle of refraction
1/1.33 = sin 25/sin R
Make sin R the subject of formula
Sin R = sin 25 / 0.7518
Sin R = 0.5620
R = sin^-1(0.5620)
R = 34.2 degrees
This angle will be an incident angle when the light is travelling from the water to glass. Using Snell law agains,
1.33/1.55 = sin 34.2/sin R
Cross multiply
1.33 sin R = 0.87122
Make R the subject of formula
Sin R = 0.87122/1.33
Sin R = 0.6550
R = sin^-1(0.6550)
R = 40.9 degrees
Therefore, the angle of refraction in the water is 34.2 degrees and then in the glass is 41 degrees approximately.
Find the attached file for the figure
To solve this problem we will apply the principles of conservation of energy, for which we have to preserve the initial kinetic energy as elastic potential energy at the end of the movement. If said equality is maintained then we can affirm that,
Here,
m = mass
k = Spring constant
x = Displacement
v = Velocity
Rearranging to find the velocity,
Our values are,
Replacing our values we have,
Therefore the velocity is
Answer:
F=248.5W N
Explanation:
Newton's 2nd Law tells us that F=ma. We will use their averages always. The average acceleration the tennis ball experimented is, by definition:
Since we start counting at 0s and the ball departs from rest, this is just
So we can write:
Where in the last step we have just multiplied and divided by g, the acceleration of gravity. This allows us to introduce the weight of the ball W since W=gm, so we have:
Substituting our values:
Where the average force exerted has been written it terms of the tennis ball's weight W.
Answer:
Theta1 = 12° and theta2 = 168°
The solution procedure can be found in the attachment below.
Explanation:
The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).
In the solution as can be found below, the expression to calculate the range for any launch angle theta was first derived and then the required angles calculated from the equation by substituting the values of the the given quantities.
