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Ksivusya [100]
2 years ago
12

An object of height 8.50 cm is placed 20.0 cm to the left of a converging lens with a focal length of 12.0 cm. Determine the ima

ge location in cm, the magnification, and the image height in cm.
Physics
1 answer:
ddd [48]2 years ago
5 0

The image distance is 33.3 cm while the image height is 14.2 cm.

<h3>What is a converging lens?</h3>

A converging lens will always have a positive focal length hence, we have to find the object distance as follows;

1/f = 1/v + 1/u

1/12 = 1/v + 1/20

1/v = 1/12 - 1/20

1/v = 0.08 - 0.05

v =33.3 cm

Now;

Magnification =  33.3 cm/20.0 cm =1.67

M = Image height/Object height

1.67 =  Image height/8.50 cm

Image height = 1.67  * 8.50 cm

Image height = 14.2 cm

Learn more about focal length:brainly.com/question/16188698

#SPJ1

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7 0
3 years ago
A professor, with dumbbells in his hands and holding his arms out, is spinning on a turntable with an angular velocity. What hap
Olenka [21]

Answer:

<em>His angular velocity will increase.</em>

Explanation:

According to the conservation of rotational momentum, the initial angular momentum of a system must be equal to the final angular momentum of the system.

The angular momentum of a system = I'ω'

where

I' is the initial rotational inertia

ω' is the initial angular velocity

the rotational inertia = mr'^{2}

where m is the mass of the system

and r' is the initial radius of rotation

Note that the professor does not change his position about the axis of rotation, so we are working relative to the dumbbells.

we can see that with the mass of the dumbbells remaining constant, if we reduce the radius of rotation of the dumbbells to r, the rotational inertia will reduce to I.

From

I'ω' = Iω

since I is now reduced, ω will be greater than ω'

therefore, the angular velocity increases.

5 0
3 years ago
In the simulation, there are three balls on the floor. Drag each of them up off the floor, and then let go. See what happens to
Vlad1618 [11]

Answer:

I hope this helps and I'm not to late

A way the balls behave the same way is by bouncing about 1 time after throwing the balls up. A way the balls act differently is the blue ball is bouncier than all the balls, the red ball bounces about 2 times before stopping, and the green ball doesn’t really bounce except for one time.

Explanation:

you also can use paraphrase to help you reword bye bye!!

7 0
2 years ago
What is the ratios of sodium hydroxide
Katena32 [7]
Clarify what you mean by ratios?
8 0
3 years ago
The next four questions refer to the situation below.
Anna11 [10]

Answer:

 t_{out} = \frac{v_s - v_r}{v_s+v_r} t_{in},      t_{out} = \frac{D}{v_s +v_r}

Explanation:

This in a relative velocity exercise in one dimension,

let's start with the swimmer going downstream

its speed is

         v_{sg 1} = v_{sr} + v_{rg}

The subscripts are s for the swimmer, r for the river and g for the Earth

with the velocity constant we can use the relations of uniform motion

           v_{sg1} = D / t_{out}

           D = v_{sg1}  t_{out}

now let's analyze when the swimmer turns around and returns to the starting point

        v_{sg 2} =  v_{sr}  - v_{rg}

         v_{sg 2} = D / t_{in}

         D = v_{sg 2}  t_{in}

with the distance is the same we can equalize

           v_{sg1} t_{out} = v_{sg2} t_{in}

          t_{out} =  t_{in}

           t_{out} = \frac{v_s - v_r}{v_s+v_r} t_{in}

This must be the answer since the return time is known. If you want to delete this time

            t_{in}= D / v_{sg2}

we substitute

            t_{out} = \frac{v_s - v_r}{v_s+v_r} ()

            t_{out} = \frac{D}{v_s +v_r}

7 0
2 years ago
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