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3 years ago

Why do concave lenses always form virtual images?

Physics

2 answers:

kolbaska11 [484]3 years ago

5 0

Answer:

Explanation:

There are two types of images formed by the lenses or mirrors.

1. Real image: When two refracted or reflected rays actually meet at a point after refraction or reflection, the image formed is real.

2. Virtual image: When two refracted or reflected rays appear to meet at a point after refraction or reflection, the image formed is virtual.

A concave lens is a diverging lens, it means when the rays of light incident on the concave lens, then after refraction, the rays diverges from its original path. Thus, they cannot meet at a point, they appears to comes from a point and the image formed is virtual.

MaRussiya [10]3 years ago

3 0

A concave mirror and a converging lens will only produce a real image if the object is located beyond the focal point (i.e., more than one focal length away). The image of an object is found to be upright and reduced in size.

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makvit [3.9K]

Use the eq. of Young modulus Y=(F/A)/(∆l/lo)
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3 years ago

Leoni is participating in four drawing competitions. If the probability of her losing any

Alex17521 [72]

Answer:

(a) $Probability = 0.7599$

(b) $Probability = 0.2646$

Explanation:

Represent losing with L and winning with W.

So:

$L = 0.7$ --- Given

$n = 4$

Probability of winning would be:

$W = 1 - L$

$W = 1 - 0.7$

$W = 0.3$

The question illustrates binomial probability and will be solved using the following binomial expansion;

$(L + W)^4 = L^4 + 4L^3W + 6L^2W^2 + 4LW^3 + W^4$

So:

Solving (a): Winning at least 1

We look at the above and we list out the terms where the powers of W is at least 1; i.e., 1,2,3 and 4

So, we have:

$Probability = 4L^3W + 6L^2W^2 + 4LW^3 + W^4$

Substitute value for W and L

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$Probability = 0.7599$

Hence, the probability of her winning at least one is 0.7599

Solving (a): Wining exactly 2

We look at the above and we list out the terms where the powers of W is exactly 2

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Substitute value for W and L

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$Probability = 0.2646$

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6 0

2 years ago

A curve that has a radius of 90 m is banked at an angle of =10.8∘. If a 1100 kg car navigates the curve at 75 km/h without skidd

PilotLPTM [1.2K]

The minimum coefficient of static friction between the pavement and the tires is 0.69.

The given parameters;

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mass of the car, m = 1100 kg

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$F_n = mgcos(\theta)$

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