Answer:
The intensity is 
Explanation:
From the question we are told that
The intensity of the unpolarized light is 
The angle between the ideal polarizing sheet is 
Generally the intensity of light emerging from the first polarizer is mathematically represented as
substituting values
Then the intensity of incident light emerging from the second polarizer is mathematically represented by Malus law as
substituting values
![I_2 = 2000 * [cos (24.58)]^2](https://tex.z-dn.net/?f=I_2%20%20%3D%202000%20%20%2A%20%5Bcos%20%2824.58%29%5D%5E2)
The bomb strike the ground relative to the point at 1km . B
<h3>How to determine the distance</h3>
Using the equation
h = 1/2 gt^2
500 = 1/2 * 10* t^2
500 = 5t^2
t = √500/5
t = √100
t = 10seconds
To find the distance,
Distance = velocity * time
Distance = 100 ÷ 10
Distance = 1000m = 1km
Therefore, the bomb strike the ground relative to the point at 1km . B
Learn more about projectile distance here:
brainly.com/question/15502195
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Answer:
Angle = 0.2520 radians
Explanation:
Complete question:
Sound with frequency 1220Hz leaves a room through a doorway with a width of 1.13m . At what minimum angle relative to the centerline perpendicular to the doorway will someone outside the room hear no sound? Use 344m/s for the speed of sound in air and assume that the source and listener are both far enough from the doorway for Fraunhofer diffraction to apply. You can ignore effects of reflections.
Given Data:
Speed of sound =v= 344 m/sec ;
Width of doorway =d= 1.13m ;
Frequency of sound =f= 1220 Hz ;
Solution:
As we know that
Wvelength = w = v/f = 344/1220 = 0.281967m
Now we also know that
w = dsin(A) where A is the angle
A = arcsin(w/d) =14.44° = 14.44*(3.14/180) = 0.2520 radians
At the angle of 0.252 radians relative to the centreline perpendicular to the doorway a person outside the room will hear no sound under given conditions.
Answer:
No more information is needed
Explanation:
Radio waves are electromagnetic energy, lower frequency forms of this type of energy that includes light and cosmic rays on the high frequency end that we are able to detect. So in free space (vacuum), radio waves travel at their fastest velocity, the “speed of light”. The reason for the quotation marks is because when light or radio waves are propagating through matter, we observe them traveling more slowly.
