PHYSICS MIDDLE SCHOOL

The force of repulsion between two identical positive charges is 0.800 N when the charges are 0.100 m apart. Find the value of each charge.

Answers

Answer 1

Answer:

Answer:

Value of each charge is given as

$q = 9.43 * 10^(-7) C$

Explanation:

As we know that electrostatic force between two charges is given as

$F = (kq_1 q_2)/(r^2)$

here we know that

$q_1 = q_2 = q$

$F = 0.800 N$

r = 0.100 m

now we have

$F = ((9* 10^9) q^2)/((0.100)^2)$

$0.800 = 9* 10^(11) q^2$

$q = 9.43 * 10^(-7) C$

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Hard well organized solid Atoms are tightly packed together
Crystalline arrangement
High melting and boiling point.

The characteristics listed describe

A) ionic compounds.
B) covalent compounds.
C) diatomic molecules.
D) molecular compounds.

Answers

A) ionic compounds.

quizlet has a great study set for this unit

https://quizlet.com/_46l8d5

Answer:

ionic compounds

Explanation:

15 POINTS! Yes, more science... What kind of tool can you use to view atoms and molecules up close?

Answers

Generally, molecules are hard to see up close, that is why models are made. So that people can see the structure of atoms and how the nucleus looks.

Other then a model, an transmissional electron microscope is used to look at atoms up close.

Well, its a microscope pretty much. But they are called "electron microscopes". But yea just a microscope. ...... oh yeah also a scanning tunneling microscope :)

Modern dual-processing theory is the idea that our minds have —tracks.
O
A. conscious and unconscious
O
B. emotional and rational
O
C. conscious and preconscious
O
D. subconscious and conscious

Answers

Modern dual-processing theory is the idea that our minds have conscious and unconscious tracks. A

Answer:

conscious and unconscious

Explanation:

Help with this question please.

Answers

There are 4 hydrogens on the right side $(2\mathrm H_2=4\mathrm H)$ , and 2 hydrogens on the left per molecule of $\mathrm H_2$ . To get the same number of hydrogens on both sides, the coefficient should be 2.

(Then the number of oxygens will be consistent, since $2\mathrm H_2\mathrm O$ contributes 2 oxygens, and so does $\mathrm O_2$ .)

1. If an object that stands 3 centimeters high is placed 12 centimeters in front of a planemirror, how far from the mirror is the image located? Explain your reasoning.

2. An object with a height of 0.3 meter is placed at a distance of 0.4 meter from a concave
spherical mirror. An image with a height of 0.1 meter is formed in front of the mirror.
How far from the mirror is the image located?

3. When an object with a height of 0.10 meter is placed at a distance of 0.20 meter from a
convex spherical mirror, the image will appear to be 0.06 meter behind the mirror.
What's the height of the image?

4. Compare and contrast the properties of the images formed by each mirror type in the
table.

Answers

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;

$Magnification, \ m = (Image \ height )/(Object \ height ) = (Image \ distance \ from \ mirror )/(Object\ distance \ from \ mirror )$

$m = (0.1)/(0.3 ) = (Image \ distance \ from \ mirror )/(0.4 )$

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. $m = (Image \ height)/(0.10 ) = (0.06 )/(0.20 )$

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.

Answer:

1. The mirror is 12 centimeters away from the image. This is a plane mirror with a flat reflecting surface. The distance between the object and the mirror surface is equal to the distance between the mirror surface and the image.

2. hiho=siso

0.1 m0.3 m=si0.4 m

Multiply each side of this equation by 0.4.

0.4×(0.10.3=si0.4)×0.4

si=0.40.3

si = 0.133 m

3. hiho=siso

hi0.10 m=0.06 m0.02 m

Multiply each side of this equation by 0.10.

0.10×(hi0.10=0.060.20)×0.10

hi=0.0060.20

hi = 0.03 m

Image Formation

Mirror Type Appearance Placement Size

Plane Virtual Erect (Upright); Appears to have left and right reversed Behind the mirror; the distance between the mirror and the image is equal to the distance between the mirror and the object Depends on the size of the mirror and placement of the object

Concave (when the object is located a distance greater than a focal length from mirror's surface) Real Inverted In front of the mirror Smaller than the object

Concave (when object is located a distance less than the focal length of the mirror) Virtual Erect (Upright) Behind the mirror Enlarged

Convex Virtual Erect (Upright) Behind the mirror Smaller than the object

Explanation:

PENN

A particle is moving in a circular path of radius r . The displacement after half a circle would be

Answers

Answer:

Explanation:

As we know that

The displacement is the total distance measured between the initial or start and final or destination point

If particle cover half path of the circle, the displacement can easily find out by considering the distance between the start and destination point

We attached the diagram for better understanding

As per the diagram.

The displacement after half-circle is

AB = OA + OB

= r + r

= 2r

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