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arlik [135]
3 years ago
8

The law of reflection is quite useful for mirrors and other flat, shiny surfaces. (This sort of reflection is called specular re

flection). However, you've likely been told that when you look at something, you are seeing light reflected from the object that you are looking at. This is reflection of a different sort of diffuse reflection.
Suppose that the spotlight shines so that different parts of the beam reflect off of different two surfaces, one inclined at an angle alpha (from the horizontal) and one inclined at an angle beta. What would the angular separation between the rays reflected from the two surfaces?
Physics
1 answer:
Sever21 [200]3 years ago
5 0

Answer:

Explanation:

Suppose initially the plane was horizontal and light was reflected back at some angle θ from the normal .

Now the reflecting surface is twisted so that is becomes inclined at angle alpha .

The reflected light will be deviated from its original direction by angle

2 x alpha .

Similarly when the reflecting surface is further twisted so that it becomes inclined at angle beta then again the reflected beam will deviated by angle

2 x beta

Hence angle between these two reflected beam

= 2 beta - 2 alpha

= 2 ( β - α )

So, angular separation between the rays reflected from the two surfaces

= 2 ( β - α ) .

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A 1.5-kilogram cart initially moves at 2.0 meters
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             Acceleration = (change in speed) / (time for the change)

Change in speed = (speed at the end) minus (speed at the beginning.

The cart's acceleration is

                               (0 - 2 m/s) / (0.3 sec)

                           = ( -2 / 0.3 ) (m/s²)  =  -(6 and 2/3) m/s² .

Newton's second law of motion says

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For this cart:      Force = (1.5 kg) x ( - 6-2/3 m/s²)

                                       = ( - 1.5 x 20/3 ) (kg-m/s²)

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7 0
3 years ago
ANSWER FAST PLZZ!!
Dahasolnce [82]

Answer:

See the answer below

Explanation:

The two correct options are:

1. <em>The student traveled the fastest between 4 and 5 minutes</em>

<em>2. The average speed of the student is 4 meters per minute</em>

Between 4 and 5 minutes, the approximate distance of 7 meters covered by the student is the longest of all the distances covered per minute. Since speed is a measure of distance covered relative to time, <u>it means that the student traveled the fastest between these periods.</u>

<em>The average speed is calculated by dividing the total distance traveled by a body by the total time taken to cover the distance. </em>

From the graph:

Total distance covered = 20

Total time taken = 5

Average speed = 20/5 = 4 meters per minute

<em>All the remaining conclusions are incorrect or indeterminate going by the information on the graph.</em>

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Look at the graph below. Which type of graph is displayed? line bar scatterplot histogram
Leona [35]

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3 years ago
Read 2 more answers
1. A small block of mass m slides in a vertical circle of radius R on the inside of a circular track. There is no friction betwe
Nookie1986 [14]

Answer:

a) v_{b} = \sqrt{\frac{R}{m}( n_{b} - mg)}

b) v_{t} = \sqrt{\frac{R}{m}( n_{b} - mg) - 4gR}

c) N_{t} = = n_{b} - 6mg

Explanation:

NOTE: No values are provided in this exercise, any values of the given parameters can be substituted into the answers provided

Let the mass of the block = m

Radius  = R

Normal force acting on the block, N =  n_{b}

a) Calculate the velocity of the block at the bottom of its path, v_{b}

According to Newton's second law of motion,

\sum F = ma\\\sum F = N - mg

N - mg = ma

Since the block clides in a vertical circle, the acceleration, a, is a centripetal acceleration

a = \frac{v_{b} ^{2} }{R}

n_{b} - mg = \frac{mv_{b} ^{2} }{R}

\frac{R}{m}( n_{b} - mg) = v_{b} ^{2} \\ v_{b} = \sqrt{\frac{R}{m}( n_{b} - mg)}

b) Use conservation of energy to find the velocity of the block at the top of its path.

According to the law of conservation of energy

K_{b} = K_{t} + PE_{t}.................(1)

PE_{t} = mgh

The height of the block's path at the top, h = twice the radius

h = 2R

PE_{t} = mg(2R)\\PE_{t} = 2mgR

From equation (1)

\frac{1}{2} mv_{b} ^{2} = \frac{1}{2} mv_{t} ^{2} + 2mgR\\\frac{1}{2} v_{b} ^{2} = \frac{1}{2} v_{t} ^{2} + 2gR\\\frac{1}{2} v_{b} ^{2} -  2gR= \frac{1}{2} v_{t} ^{2}

Substituting v_{b} into the equation above

\frac{1}{2} (\sqrt{\frac{R}{m}( n_{b} - mg)})^{2} - 2gR = \frac{1}{2} v_{t} ^{2} \\\frac{R}{m}( n_{b} - mg) - 4gR = v_{t} ^{2}\\v_{t} = \sqrt{\frac{R}{m}( n_{b} - mg) - 4gR}

c) Find the normal force that the track exerts on the block at the top of its path.

From the Newton's law of motion applied in part a

N_{t} + mg = ma

a = \frac{v_{t} ^{2} }{R}

N_{t} = \frac{mv_{t}^{2}} {R} - mg

N_{t} = \frac{m(\frac{R}{m}( n_{b} - mg) - 4gR) }{R} - mg\\N_{t} = (n_{b} - mg) - 4mg - mg\\N_{t} = = n_{b} - 6mg

6 0
3 years ago
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