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Savatey [412]
3 years ago
13

1. If an object that stands 3 centimeters high is placed 12 centimeters in front of a plane

Physics
1 answer:
igor_vitrenko [27]3 years ago
3 0

Answer:

1. 12 cm

2. 0.133 m

3. 0.03 m

4. Plane mirror

Virtual image

Upright

Behind the mirror

The same size as the object

Concave mirror when the object is located a distance greater than the focal length from the mirror's surface

Real image

Inverted image

In front of the the mirror

Diminished when the object is beyond the center of curvature

Same size as object when the object is placed at the center of curvature

Enlarged when the object is placed between the center of curvature of the mirror and the focus of the mirror

Concave mirror when the object is located a distance less than the focal length from the mirror's surface

Virtual image

Upright image

Behind the the mirror

Enlarged

Convex mirror

Type = Virtual image

Appearance = Upright image

Placement = Behind the mirror

Size = Smaller than the object

Explanation:

1. For plane mirror, since there is no magnification, the virtual image distance from the mirror = object distance from the mirror = 12 cm behind the mirror

2. The height of the object = 0.3 m

The distance of the object from the mirror = 0.4 meters

Height of image formed = 0.1 meter

We have;

Magnification, \ m = \dfrac{Image \ height }{Object \ height } = \dfrac{Image \ distance \ from \ mirror }{Object\ distance \ from \ mirror }

m = \dfrac{0.1}{0.3 } = \dfrac{Image \ distance \ from \ mirror }{0.4 }

Image distance from the mirror = 0.1/0.3×0.4 = 2/15 = 0.133 m

Image distance from the mirror = 0.133 m

3. m = \dfrac{Image \ height}{0.10 } = \dfrac{0.06 }{0.20 }

The image height = 0.06/0.2×0.1 = 3/100 = 0.03 meter

The image height = 0.03 meter

4. Plane mirror

Type = Virtual image

Appearance = Upright image with the left transformed to right

Placement = Behind the mirror

Size = The same size as the object

Concave mirror when the object is located a distance greater than the focal length from the mirror's surface

Type = Real image

Appearance = Inverted image

Placement = In front of the the mirror

Size = Diminished when the object is beyond the center of curvature

Same size as object when the object is placed at the center of curvature

Enlarged when the object is placed between the center of curvature of the mirror and the focus of the mirror

Concave mirror when the object is located a distance less than the focal length from the mirror's surface

Type = Virtual image

Appearance = Upright image

Placement = Behind the the mirror

Size = Enlarged

Convex mirror

Type = Virtual image

Appearance = Upright image

Placement = Behind the mirror

Size = Smaller than the object.

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At baseball practice, Mason and Alfredo both picked up the same bat and neither would let go until one of them had it for himsel
Illusion [34]

Answer:

Option B. O because the net force was 5 N in Alfredo's direction

Explanation:

To know the the correct answer to the question given above, we shall determine the net force acting on the bat. This can be obtained as follow:

Force of pull by Mason (Fₘ) = 15 N

Force of pull by Alfredo (Fₐ) = 20 N

Net force (Fₙ) =?

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Fₙ = 5 N in Alfredo's direction

From the calculation made above, we can see that the net force is 5N in Alfredo's direction. This is the reason why Alfredo end up having the bat.

7 0
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8 0
3 years ago
What is the relationship between an object’s mass and its gravitational potential energy?.
antiseptic1488 [7]

Answer:

If we’re talking about objects on the Earth, the gravitational potential energy is given by:

Explanation:

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5 0
2 years ago
Alcohol of mass 33.2g and density 0.79kg/m³ or 790kg/m³ is mixed with water of 9g. What is the density of the resulting mixture?
KATRIN_1 [288]

Answer:

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Explanation:

The volume of the alcohol is ...

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The density of water is about 1 g/cm³, so the volume of 9 g of it is ...

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So, the resulting density is ...

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The resulting mixture has a density of about 0.83 g/cm³.

_____

<em>Additional comment</em>

Alcohol dissolves in water, so the total volume will be slightly less than 51.0253 cm³. The attached curve shows the result of mixing ethanol and water.

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