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brilliants [131]
4 years ago
10

A large convex lens stands on the floor. The lens is 180 cm tall, so the principal axis is 90 cm above the floor. A student hold

s a flashlight 120 cm off the ground, shining straight ahead (parallel to the floor) and passing through the lens. The light is bent and intersects the principal axis 60 cm behind the lens. Then the student moves the flashlight 30 cm higher (now 150 cm off the ground), also shining straight ahead through the lens. How far away from the lens will the light intersect the principal axis now?
Physics
1 answer:
gayaneshka [121]4 years ago
5 0
B. 60 cm 

  All parallel light rays are bent through the focal point of a convex lens, so the rays from the flashlight 150 cm above the floor must go through the same point on the principal axis as the rays from the flashlight 120 cm above the floor. The location of the focal point does not change when the position of the object is moved either vertically or horizontally.
Hope this helps !
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An object is in uniform circular motion, tracing an angel at 30 degrees every 0.010 seconds. What's the period of this motion an
Neko [114]
Here's the rule you need to know
in order to answer this question:

                     1 full circle ==> 360 degrees .

Got that ?

Now you could set up a proportion:

     (30 degrees) / (0.01 second)  =  (360 degrees) / (time for full period)

Cross-multiply the proportion:

     (30°) · (period)  =  (360°) · (0.01 sec)

Divide each side by (30°) :    Period = (360° · 0.01 sec) / (30°)

                                                     =  (3.6° · sec) / (30°)

                                                     =  (3.6 / 30)  sec

                                                     =      0.12  sec .
___________________________________

Another way to look at it:

30°        takes    0.01 second
60°        takes    0.02 second
90°        takes    0.03 second
120°      takes    0.04 second
150°      takes    0.05 second
180°      takes    0.06 second
210°      takes    0.07 second
240°      takes    0.08 second
270°      takes    0.09 second
300°      takes    0.10 second
330°      takes    0.11 second
360°      takes   0.12 second

7 0
4 years ago
The potential-energy function u(x) is zero in the interval 0≤x≤l and has the constant value u0 everywhere outside this interval.
VMariaS [17]
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So λ is proportional to 1/√K. 
in the potential well the potential energy is zero, so completely the electron's energy is in the shape of kinetic energy: 
K = 6U₀ 

Outer the potential well the potential energy is U₀, so 
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Therefore, the ratio of the de Broglie wavelength of the electron in the region x>L (outside the well) to the wavelength for 0<x<L (inside the well) is: 
1/√(5U₀) : 1/√(6U₀) 
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5 0
4 years ago
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ivanzaharov [21]

Answer:

825 m

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I would say the same thing as the first answer
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