Answer:
a) 46.5º b) 64.4º
Explanation:
To solve this problem we will use the laws of geometric optics
a) For this part we will use the law of reflection that states that the reflected and incident angle are equal
θ = 43.5º
This angle measured from the surface is
θ_r = 90 -43.5
θ_s = 46.5º
b) In this part the law of refraction must be used
n₁ sin θ₁ = n₂. Sin θ₂
sin θ₂ = n₁ / n₂ sin θ₁
The index of air refraction is n₁ = 1
The angle is this equation is measured between the vertical line called normal, if the angles are measured with respect to the surface
θ_s = 90 - θ
θ_s = 90- 43.5
θ_s = 46.5º
sin θ₂ = 1 / 1.68 sin 46.5
sin θ₂ = 0.4318
θ₂ = 25.6º
The angle with respect to the surface is
θ₂_s = 90 - 25.6
θ₂_s = 64.4º
measured in the fourth quadrant
Answer:
Static Friction - acts on objects when they are resting on a surface
Sliding Friction - friction that acts on objects when they are sliding over a surface
Rolling Friction - friction that acts on objects when they are rolling over a surface
Fluid Friction - friction that acts on objects that are moving through a fluid
Explanation:
Examples of static include papers on a tabletop, towel hanging on a rack, bookmark in a book
, car parked on a hill.
Example of sliding include sledding, pushing an object across a surface, rubbing one's hands together, a car sliding on ice.
Examples of rolling include truck tires, ball bearings, bike wheels, and car tires.
Examples of fluid include water pushing against a swimmer's body as they move through it , the movement of your coffee as you stir it with a spoon, sucking water through a straw, submarine moving through water.
Answer:
alpha=53.56rad/s
a=5784rad/s^2
Explanation:
First of all, we have to compute the time in which point D has a velocity of v=23ft/s (v0=0ft/s)
Now, we can calculate the angular acceleration (w0=0rad/s)
with this value we can compute the angular velocity
and the tangential velocity of point B, and then the acceleration of point B:
hope this helps!!
Answer:
Leak 1 = 3.43 m/s
Leak 2 = 2.42 m/s
Explanation:
Given that the top of the boot is 0.3 m higher than the leaks.
Let height H = 0.3m and the acceleration due to gravity g = 9.8 m/s^2
From the figure, the angle of the leak 1 will be approximately equal to 45 degrees. While the leak two can be at 90 degrees.
Using the third equation of motion under gravity, we can calculate the velocity of leak 1 and 2
Find the attached files for the solution and figure
The motion of a falling whirligig is different to that of a falling paper ball due to spinning.
<h3>Type of motion performed by whirligig and falling paper ball </h3>
The motion of a falling whirligig is different from the motion of a falling paper ball because the paper ball falls on the ground without spinning while on the other hand, the whirligig falls on the ground along with spinning.
The falling whirligig performs two motion i.e. one is falling on the ground and the other is spinning during motion whereas paper ball performs one motion i.e. motion in the air towards the ground so we can conclude that the motion of a falling whirligig is different than of a falling paper ball.
Learn more about motion here: brainly.com/question/453639
