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Lemur [1.5K]
3 years ago
6

For a plane mirror,how is the object s related to the distance s?

Physics
1 answer:
dlinn [17]3 years ago
3 0
Light is a very complex phenomenon, but in many situations its behavior can be understood with a simple model based on rays and wave fronts. A ray is a thin beam of light that travels in a straight line. A wave front is the line (not necessarily straight) or surface connecting all the light that left a source at the same time. For a source like the Sun, rays radiate out in all directions; the wave fronts are spheres centered on the Sun. If the source is a long way away, the wave fronts can be treated as parallel lines.

Rays and wave fronts can generally be used to represent light when the light is interacting with objects that are much larger than the wavelength of light, which is about 500 nm. In particular, we'll use rays and wave fronts to analyze how light interacts with mirrors and lenses.

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A rocket ship starts from rest and turns on its forward booster rockets causing it to have a constant acceleration of 4 meters p
bagirrra123 [75]

Complete question is;

A rocket ship starts from rest and turns on its forward booster rockets, causing it to have a constant acceleration of 4 m/s² rightward. After 3s, what will be the velocity of the rocket ship?

Answer:

v = 12 m/s

Explanation:

We are given;

Initial velocity; u = 0 m/s (because ship starts from rest)

Acceleration; a = 4 m/s²

Time; t = 3 s

To find velocity after 3 s, we will use Newton's first equation of motion;

v = u + at

v = 0 + (4 × 3)

v = 12 m/s

6 0
3 years ago
What will be the final velocity of a rock if we drop it off of a bridge and it strikes the ground 2.8s later (ignoring air resis
mars1129 [50]
Formula for final velocity: Vf= vi+(a*t)
Vi- initial velocity, a=acceleration, t-time

Vf=vi+(at)
Vf= 0+(9.8m/s*2.8s)
Vf= 27.44 m/s

The acceleration of the Earth when dropping something would be 9.8 m/s

Here is an reference that can help you answer problems like these.
Hope this helps and good luck :)

5 0
2 years ago
a 2.5 kg rock is dropped off a 32 m cliff and hits a spring, compressing it 57cm. what is the spring constant
Rama09 [41]

2.5 kg because you cant change the weight of the rock

4 0
2 years ago
Read 2 more answers
What is the formula for the moment of inertia of the person/single particle rotating in a circle? (Give these values with a subs
Ann [662]

Moment of inertia of single particle rotating in circle is I1 = 1/2 (m*r^2)

The value of the moment of inertia when the person is on the edge of the merry-go-round is I2=1/3 (m*L^2)

Moment of Inertia refers to:

  • the quantity expressed by the body resisting angular acceleration.
  • It the sum of the product of the mass of every particle with its square of a distance from the axis of rotation.

The moment of inertia of single particle rotating in a circle I1 = 1/2 (m*r^2)

here We note that the,

In the formula, r being the distance from the point particle to the axis of rotation and m being the mass of disk.

The value of the moment of inertia when the person is on the edge of the merry-go-round is determined with parallel-axis theorem:

I(edge) = I (center of mass) + md^2

d be the distance from an axis through the object’s center of mass to a new axis.

I2(edge) = 1/3 (m*L^2)

learn more about moment of Inertia here:

<u>brainly.com/question/14226368</u>

#SPJ4

7 0
2 years ago
A student wants to start a small business in school. Write down six items that
Fiesta28 [93]

Answer:

packets of pen

packets of pencil

copies

books

bottles

mask

3 0
2 years ago
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