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Wewaii [24]
1 year ago
14

an object is placed at 19.3 cm in front of a diverging lens with a focal length of -18.7 cm. what is the magnification? enter a

number with 2 digits behind the decimal point.
Physics
1 answer:
Bezzdna [24]1 year ago
8 0

The magnification <u>is 31.16.</u>

Magnification is the process of increasing the apparent size of something rather than its physical size. This increase is quantified by a calculated number, also called the "factor". If this number is less than 1, it means size reduction, sometimes called size reduction or reduction.

u =  -19.3

f = -18.7 cm.

m = f/f-u

    = -18.7/-18.7 +19.3

   <u>= 31.16</u>

The term magnification refers to the size of the image produced by the lens compared to the size of the object. For lenses: Magnification "m" is the ratio of image height to object height. The magnification of a lens is defined as the ratio of image height to object height. It is also given by image distance and object distance. equal to the ratio of image distance to object distance.

Learn more about magnification here:-brainly.com/question/15744335

#SPJ4

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Answer:

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Explanation:

1.

1 x 10⁻⁶ s = 1 μs

(1 x 10⁻⁶ x 10⁶) s = 1 x 10⁶ μs

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2.

1000 g = 1 kg

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3.

<u>1 km = 1000 m</u>

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4.

<u>1 mm = 1 x 10⁻³ m</u>

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5.

<u>1 mL = 1 x 10⁻³ L</u>

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6.

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A disk of a radius 50 cm rotates at a constant rate of 100 rpm. What distance in meters will a point on the outside rim travel d
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Answer:

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Explanation:

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The distance in meters traveled by a point outside the rim is calculated as follows;

\theta = \omega t\\\\\theta = (100 \frac{rev}{\min}  \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 s} ) \times (30 s)\\\\\theta = 100 \pi \ rad\\\\d = \theta r\\\\d = 100\pi  \ \times \ 0.5m\\\\d = 50 \pi \ m = 157.1 \ m

Therefore, the distance in meters traveled by a point outside the rim is 157.1 m

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