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Pani-rosa [81]
2 years ago
10

The focal length of a diverging lens is negative. If f = −26 cm or a particular diverging lens, where will the image be formed o

f an object located 63 cm to the left of the lens on the optical axis?
.... cm to the left of the lens

....What is the magnification of the image?
Physics
1 answer:
Arada [10]2 years ago
7 0

The distance of the image formed is 44.27 cm and the magnification of the image is 0.7.

<h3>Location of the image</h3>

The image distance formed by the lens is calculated as follows;

1/v + 1/u = -1/f

where;

  • v is the image distance
  • u is the object distance
  • f is the focal length of the lens

1/v = -1/f - 1/u

1/v = -(-1/26) - 1/63

1/v = 1/26 - 1/63

1/v = 0.022588

v = 1/0.022588

v = 44.27 cm

<h3>What is magnification of lens?</h3>

The magnification of a lens is defined as the ratio of the height of an image to the height of an object.

<h3>Magnification of the image formed</h3>

The magnification of the image is calculated as follows;

Magnification = image distance/object distance

M = 44.27/63

M = 0.7

Thus, the distance of the image formed is 44.27 cm and the magnification of the image is 0.7.

Learn more about magnification here: brainly.com/question/1599771

#SPJ1

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A body moves in a straight line. At time, it's acceleration is given by a = 12t + 1. When t =0, the velocity of the body v is 2
madreJ [45]

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59 m/s, 64.5 m

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At t = 0, v = 2.

2 = 6(0)² + (0) + C

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s = ∫ v dt

s = 2t³ + ½ t² + 2t + C

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0 = 2(0)³ + ½ (0)² + 2(0) + C

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At t = 3:

v = 6(3)² + (3) + 2 = 59

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A 350-g mass is attached to a spring whose spring constant is 64 N/m. Its maximum acceleration is 5.3 m/s2. What is the frequenc
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The frequency of oscillation is 2.153 Hz

What is the frequency of spring?

Spring Frequency is the natural frequency of spring with a weight at the lower end. Spring is fixed from the upper end and the lower end is free.

For the mass-spring system in this problem,

The Frequency of spring is calculated with the equation:

f = \frac{1}{2\pi } \sqrt{\frac{k}{m} }

Where,

f = frequency of spring

k = spring constant = 64 N/m

m = mass attached to spring = 350g = 0.350 kg

a = maximum acceleration = 5.3 m/s^2

Substituting the values in the equation,

f = \frac{1}{2\pi } \sqrt{\frac{64}{0.350} }

f = \frac{1}{2\pi } ( 13.522)

f = 2.1535 Hz

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The frequency of oscillation is 2.153 Hz

Learn more about frequency here:

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