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1 year ago

8

Three flat layers of transparent material are stacked upon one another. The top layer has index of refraction n1, the middle has

n2 and the bottom one has n3. If n1 > n2 > n3, and if a ray of light strikes the top layer at an angle of incidence, in which layer is the angle of refraction the greatest? Why?
a. the bottom layer
b. the top layer
c. Once the ray enters the touching layers, the angle of refraction remains constant.
d. the middle layer

1 answer:

serg [7]1 year ago

7 0

Answer:

a. the bottom medium

Explanation:

it has the least index of refraction and hence most rarer.

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To visit your favorite ice cream shop, you must travel 490 m west on Main Street and then 920 m south on Division Street. Suppos

topjm [15]

Answer:

a) The magnitude of your average velocity during the 121 s is 8.61 m/s.

b) The direction of the average velocity is 61.9° south of west.

c) Your average speed during the trip is 11.7 m/s

Explanation:

Hi there!

a) The average velocity (a.v) is calculated as the displacement divided by the time it took to do such a displacement.

The displacement is calculated as the distance between the initial position and the final position:

Displacement = Δ(x,y) = final position - initial position

Let's consider that your initial position is the origin of our frame of reference and let's also consider that west and south are positive directions (+x and +y respectively). Then the displacement vector will be:

Δ(x,y) = final positon - initial position

Δ(x,y) = (490, 920) m - (0, 0) m = (490, 920) m

The average velocity will be:

a.v = Δ(x,y) / t

a.v = (490, 920) m / 121 s

a.v = (4.05, 7.60) m/s

The magnitude of the average velocity is calculated as follows:

The magnitude of your average velocity during the 121 s is 8.61 m/s.

b) To find the direction of the average velocity, we have to use trigonometric rules of right triangles. Notice that the x and y-components of the average velocity (vx and vy) together with the average velocity vector (v), with magnitude 8.61 m/s, form a triangle (see figure).

Also, notice that v is the hypotenuse of the triangle and that vx is the side adjacent to the angle θ while vy is the side opposite to θ.

Using trigonometry, we can calculate the value of the angle θ:

cos θ = adjacent side / hypotenuse

cos θ = vx / v

cos θ = 4.05 m/s / 8.61 m/s

θ = 61.9°

The direction of the average velocity is 61.9° south of west.

c) The average speed (a.s) is calculated as the traveled distance (d) divided by the time it took to cover that distance (t). In total, you traveled (490 m + 920 m) 1410 m in 121 s, then the average speed will be:

a.s = d/t

a.s = 1410 m / 121 s

a.s = 11.7 m/s

Your average speed during the trip is 11.7 m/s

5 0

2 years ago

A penny falls from a windowsill which is 25.0 m above the sidewalk. How much time does a passerby on the sidewalk below have to

Yakvenalex [24]

T = √(h)/(0.5)(9.81)
t = √(25)/(4.905)
t = √5.1
t = 2.26 seconds

hope this helps and have a great day :)

3 0

2 years ago

An object’s velocity is measured to be vx(t) = α - βt2, where α = 4.00 m/s and β = 2.00 m/s3. At t = 0 the object is at x = 0. (

Leona [35]

Answer:

Explanation:

Given

$v_x(t)=\alpha -\beta t^2$

$\alpha =4\ m/s$

$\beta =2\ m/s^3$

$v_x(t)=4-2t^2$

$v=\frac{\mathrm{d} x}{\mathrm{d} t}$

$\int dx=\int \left ( 4-2t^2\right )dt$

$x=4t-\frac{2}{3}t^3$

acceleration of object

$a=\frac{\mathrm{d} v}{\mathrm{d} t}$

$a=-4t$

(b)For maximum positive displacement velocity must be zero at that instant

i.e. $v=0$

$4-2t^2=0$

$t=\pm \sqrt{2}$

substitute the value of t

$x=4\times \sqrt{2}-\frac{2}{3}\times 2\sqrt{2}$

$x=3.77\ m$

7 0

1 year ago

A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so tha

Dvinal [7]

Answer:

0.087 m

Explanation:

Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

time period, T = 3 s

the formula for the time period of the pendulum is given by

$T = 2\pi \sqrt{\frac{I}{mgd}}$ .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

Moment of inertia of the rod about the centre of mass, Ic = mL²/12

By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

I = Ic + md²

$I = \frac{mL^{2}}{12}+ md^{2}$

Substituting the values in equation (1)

$3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}$

$9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )$

12d² -26.84 d + 2.25 = 0

$d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}$

$d=\frac{26.84\pm 24.75}{24}$

d = 2.15 m , 0.087 m

d cannot be more than L/2, so the value of d is 0.087 m.

Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

3 0

1 year ago

The following represents a process used to assemble a chair with an upholstered seat. stationsa, b, and c make the seat; station

Mekhanik [1.2K]

There are three questions here:

A. The possible daily output of this process if there is 8 processing time each day?

the time it takes to assemble a chair in seconds = A + B + C + J + K + L + X + Y + Z

which is equal to = 38 + 34 + 35 + 32 + 30 + 34 + 22 + 18 + 20 =263 per chair

Processing hours = 8 hours x 60 minutes x 60 seconds

= 8 x 60 x 60

= 480 x 60 = 28,800 seconds is available in an 8 hr day.

28,800 / 263 =109.5057034220532

Therefore, It is possible to make 109 chairs in an 8 hour day.

B. Given your output rate in above, what is the efficiency of the process?

The time it takes to assemble a chair in seconds = A+ B + C + J + K + L + X + Y + Z.

= 34 + 34 + 34 + 30 + 30 + 30 + 22 + 18 + 20 = 252.6315789473684 per chair

A total of 252 per chair

Processing hours = 8 hours x 60 minutes x 60 seconds

= 8 x 60 x 60

= 480 x 60 = 28,800 seconds is available in an 8 hour day

= 28,000 / 252 = 114.2857142857143114 chairs

= 109/114 x 100 = 95.6140350877193

Therefore, the efficiency of making the chair is 95.61%

C. What is the flow time of the process?

Flow time is the period that it takes a completed chair to flow through the process from the beginning assemblage step to the last step. Take note that ABC and JKL are parallel legs in the process, and as a result both do not include to flow time to the procedure. In addition, Flow time comprises both pass time and run time at each position in the procedure.

7 0

2 years ago