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aleksley [76]
2 years ago
13

Create a ray diagram for eyeglasses that contain a diverging lens. Assume you are looking at a 2 cm tall object that is 4 cm fro

m the lens. The focal length is 3 cm.
Physics
1 answer:
k0ka [10]2 years ago
8 0

The ray diagram for the given object consists of 2 cm height of object, 4 cm object distance and 3 cm focal length.

<h3>Image formed by a diverging lens</h3>

Diverging lens is called a concave lens. The working of the lens is dependent on the refraction of the light rays as they pass through the lens.

Image formed by a diverging lens is always virtual, erect and diminished; smaller than the object and is located on the same side of the lens as the object.

The ray diagram for the given object is presented in the image in the diagram.

  • Object height = 2 cm
  • Focal length = 3 cm
  • Object distance = 4 cm

Learn more about diverging lens here: brainly.com/question/3140453

#SPJ1

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1) A boy drags a wooden crate with a mass of 20 kg, a distance of 12 m, across a rough level floor at a constant speed of 1.5 m/
mojhsa [17]

Answer: a) 49.560 and 21.13 b) i) 50 N, ii) 196 N iii) 196 N iv) 47.685 N

c) i) 594.72 ii) 0 iii) 0 iv) 0

d) 594.72

Explanation: question a)

The force is inclined at an angle of 25° to the horizontal

The horizontal component of force = 50 cos 25° = 49.560 N

The vertical component of force = 50 sin 30°= 21.130N

Question b)

i) according to the question applied force is 50 N

ii) if g = 9.8m/s², w=mg where m = mass of object = 20kg hence weight = 20* 9.8 = 196 N

iii) the normal force is the force the floor exerts on the body as a result of the weight of the object.

Normal reaction R = W = mg, we already deduced that w = mg, hence R = 196 N.

iv) according to newton's laws of motion

F - Fr = ma

F = applied force = horizontal component of force = 49.560 N.

We need to get the acceleration (a) by using Newton laws of motion before we can be able to compute the frictional force..

The body started from rest hence initial velocity u = 0

Final velocity v = 1.5m/s distance covered (s) = 12m

v ² = u² + 2as

But u = 0

v² = 2as

1.5² = 2(a) * 12

2.25 = 24a

a = 2.25/24 = 0.09735m/s²

From F - Fr = ma

49.560 - Fr = 20 * 0.09735

49.560 - Fr = 1.875

Fr = 49.560 - 1.875

Fr = 47.685 N

Question c)

i) The applied force = 49.560 N, distance covered = 12m

Work done = force * distance

Work done = 49.560 * 12

Work done = 594.72 J

ii) the weight of the object does not make the object move a distance, hence work done = 0 ( since distance covered is 0)

iii) the normal force is the same thing as the weight and they did not cover any distance hence work done is zero.

iv) the frictional force does not cover any distance, hence work done is zero.

Question d)

The total work done = work done by applied force + work done by weight + work done by normal reaction + work done by frictional force.

Total work done = 594.72 + 0 + 0 + 0 = 594.72 J

8 0
3 years ago
Across:
vladimir2022 [97]

Answer:

it would be c i just had it

Explanation:

welcome................

6 0
2 years ago
The musical tone a piano has a frequency of 346HZ and a wavelength of 1.3 M what is the speed of the sound
prohojiy [21]

Answer:

v= 449.8 m/s

Explanation:

Given data

Frequency=  346Hz

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The expression below is used to find the speed

v= f \lambda\\\\

substitute

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And so now we can calculate the tangential speed of the child, which is the angular speed times the distance of the child from the center of the motion:
v= \omega r = (0.75 rad/s)(4.65 m)=3.50 m/s
3 0
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