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Alex777 [14]
3 years ago
6

A convex mirror is placed to the right of an object. The image formed by the mirror will be a

Physics
2 answers:
Harrizon [31]3 years ago
6 0

I think it’s C but I’m not too sure

Amanda [17]3 years ago
4 0
<h3><u>Answer;</u></h3>

C.) Virtual Image that appears on the right side of the mirror.

A convex mirror is placed to the right of an object. The image formed by the mirror will be a  <u><em>Virtual Image that appears on the right side of the mirror. </em></u>

<h3><u> Explanation;</u></h3>
  • <em><u>A virtual image is an image that can not be formed on a screen. This type of an image is formed when the rays of light appear to meet at a point after reflection or refraction. Virtual images are always erect or upright.</u></em>
  • A real image on the other hand, is an image that can be formed on a screen. Real image is formed when the rays of light meet at some point after reflection or refraction. Real images are always inverted.



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A stream of electrically charged particles flowing from the sun is called the
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A basic physics question
Anna71 [15]
<h3>1)</h3>

No,kinetic energy cannot be negative since its given by KE=mv²,mass cannot be negative and the square of speed cannot b negative.

<h3>2)</h3>

Yes,any force opposing motion or displacement does negative work. They are often referred to as resistive forces (friction,air resistance,drag...)

<h3>3)</h3>

Nope it does not, it just forces the object to move in a circular path known as a centripetal force. It can accelerate an object by changing it's direction but not it's speed.

<h3>4)</h3>

No it cannot,If an object is sliding on the table (assuming it is not an incline), then most probably that normal force cancels out the weights effect or assuming there is an incline, it cancels the weight's y component.

<h3>5)</h3>

w = f.d.cos \alpha  = f.d.cos90 = 0

The work done is zero

<h3>6)</h3>

w = f.d.cos \theta = f.d.cos0 = f.d

The work is just the product of the magnitude of the force exerted and the displacement of the object.

<h3>7)</h3>

0 <  \theta < 90 \\ 1 < cos \theta < 0

<h3>Work is decreasing but positive</h3>

<h3>8)</h3>

\theta = 90 \\ cos \theta = cos90 = 0

<h3>Work is zero</h3>

<h3>9)</h3>

90 <  \theta < 180 \\ 0 < cos \theta <  - 1

<h3>Work is negative</h3>
3 0
2 years ago
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