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aleksley [76]
3 years ago
14

Which types of electromagnetic waves have frequencies that are higher than those of ultraviolet light but lower than those of ga

mma rays?
microwaves

X-rays

infrared light

radio waves
Physics
2 answers:
yKpoI14uk [10]3 years ago
7 0

The correct answer to the question is X- rays i.e the electromagnetic wave which has frequency that is higher than ultra violet wave and smaller than gamma ray is X-ray.

EXPLANATION:

As per the electromagnetic spectrum of waves, gamma ray has the least wavelength i.e it has the maximum frequency. The minimum frequency that is possessed by electromagnetic wave is radio wave.

If we arrange all the electromagnetic radiations in ascending order of frequency, the perfect order will be -

Radio wave < Micro wave < Infrared wave < Visible light < U.V ray < X - ray < Gamma ray.

From the electromagnetic spectrum, it is obvious that X -rays lie between gamma ray and U.V ray . Its frequency is smaller than gamma ray but higher  than ultra violet wave.

Luba_88 [7]3 years ago
3 0
The answer would be X-Rays.
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Issac and Blaise decide to race. They both start at the same position at the same time. Issac runs at 2m/s but decides to take a
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Let the Blaise runs for time "t" to complete the race

so the total distance he moved is given by

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3 years ago
Two converging lenses are placed 30 cm apart. The focal length of the lens on the right is 20 cm while the focal length of the l
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Answer:

a)   I2 = 3 (o-10) / (o- 30) , b)   h ’/h=  3 (o-10) / o (o-30)

Explanation:

The builder's equation is

          1 / f = 1 / o + 1 / i

Where f is the focal length, or e i are the distance to the object and image, respectively

As the separation between the lenses is greater than the focal distances, we must work them individually and separately. Let's start with the leftmost lens with focal length f = 15 cm

Let's calculate the position of the image of this lens

         1 / i1 = 1 / f - 1 / o

         1 / i1 = 1/15 - 1 / o

         i1 = o 15 / (o-15)

Let's calculate the distance to the image of the second lens, for this the image of the first is the distance to the object of the second

        o2 = d-i1

We write the builder equation

       1 / f2 = 1 / o2 + 1 / i2

       1 / i2 = 1 / f2 -1 / o2

       1 / i2 = 1 / f2 - 1 / (d-i1)

       1 / i2 = 1/20 - 1 / (d-i1)            (1)

Let's evaluate the last term

      d-i1 = d - 15 o / (o-15)

      d-i1 = (d (o-15) - 15 o) / (o-15)

      d- i1 = (30 or -30 15 -15 o) / (o-15)

      d-i1 = (15 or - 450) / (o- 15)

      d-i1 = = (15 or -450) / (o-15)

replace in 1

      1 / i2 = 1/20 - (or - 15) / (15 or -450)

      1 / i2 = [(15 o-450) - (o-15) 20] / (15 or -150)

      1 / i2 = (15 or - 450 - 20 or + 300) / (15 or - 150)

      1 / i2 = (-5 or -150) / (15 or -150)

      1 / i2 = (or -30) / (3 or - 30)

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Part B

The height of the image, we use the magnification equation

     m = h ’/ h = - i / o

     h ’= - h i / o

In our case

     h ’= h i2 / o

     h ’= h 3 (o-10) / o (o-30)

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