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Viefleur [7K]
1 year ago
11

I WILL MARK U BRAINLIEST IF U ANSWER THIS QUESTION! NEED IT ASAP PLS

Physics
1 answer:
Anuta_ua [19.1K]1 year ago
8 0
  1. One common use of a convex mirror is as shaving mirror.
  2. One common use of convex mirror is as rear-view mirrors in automobiles vehicles.

<h3>What is a concave mirror?</h3>

A concave mirror is also referred to as a converging mirror and it can be defined as a type of mirror that is designed and developed with a reflective surface that is typically curved inward and away from the source of light.

Basically, one common use of a convex mirror include the following:

  • Shaving mirrors
  • Searchlights
  • Dental mirrors.

<h3>What is a convex mirror?</h3>

A convex mirror is also referred to as a diverging mirror and it can be defined as a type of mirror that is designed and developed with a reflective surface that typically bulges outward toward the source of light.

Basically, one common use of convex mirror is as rear-view mirrors in automobiles vehicles.

Read more on convex mirror here: brainly.com/question/24175067

#SPJ1

You might be interested in
How will unbalanced forces affect the speed and direction of an object
azamat

Answer:

Explanation:

Unbalanced forces will result in the presence of acceleration. The formula

F net = ma

says that if there is a net force present and the object in question has a mass, then an acceleration is present. Now acceleration is constant in this situation because nowhere does it say the acceleration is changing. If acceleration is constant then the velocity is increasing at a steady pace (think linear function!).

The direction of the object depends on the direction that the net force is in. If the net force is to the left, then that object will accelerate to the left.

Hope this helps :)

3 0
3 years ago
A laser of wavelength 720 nm illuminates a double slit where the separation between the slits is 0.22 mm. Fringes are seen on a
kumpel [21]

Answer:

The appropriate solution is "2.78 mm".

Explanation:

Given:

\lambda = 720 \ nm

or,

  = 720\times 10^{-9} \ m

D=0.85 \ m

d = 0.22 \ mm

or,

  =0.22 \times 10^{-3} \ m

As we know,

Fringe width is:

⇒ \beta=\frac{\lambda D}{d}

hence,

Separation between second and third bright fringes will be:

⇒ \theta=\beta=\frac{\lambda D}{d}

       =\frac{720\times 10^{-9}\times 0.85}{0.22\times 10^{-3}}

       =2.78\times 10^{-3} \ m

or,

       =2.78 \ mm

8 0
2 years ago
Calculate the AMA of an access ramp if it takes 255 N of force to push a person in a wheelchair having a combined weight of 764
Andreyy89
Actual Mechanical Advantage(AMA) = Weight / Force
Here, Weight = 764 N
Force = 255 N

Substitute the values in to the expression, 
AMA = 764 / 255
AMA = 2.99

After rounding-off to the nearest tenth value, it would be 3

Finally, option C would be your answer.

Hope this helps!
7 0
3 years ago
Frank has a paperclip. It has a mass of 12g and a volume of 3cm3. What is its density?​
Mkey [24]

Answer:

D = 4 g/cm³

Explanation:

Density = Mass / Volume

Step 1: Define

D = x

M = 12 g

V = 3 cm³

Step 2: Substitute

D = 12g/3 cm³

Step 3: Simplify

D = 4g / cm³

4 0
3 years ago
Astronomers observe the motion of four planets that orbit a star similar to the Sun. Each planet follows an elliptical orbit aro
Sergeu [11.5K]

Answer:

planet that is farthest away is planet X

kepler's third law

Explanation:

For this exercise we can use Kepler's third law which is an application of Newton's second law to the case of the orbits of the planets

          T² = (\frac{4\pi ^2}{ G M_s} a³ = K_s a³

           

Let's apply this equation to our case

          a = \sqrt[3]{ \frac{T^2}{K_s} }

for this particular exercise it is not necessary to reduce the period to seconds

Plant W

             10² = K_s  a_{w}^3

             a_w = \sqrt[3]{ \frac{100}{ K_s} }

             a_w = \frac{1}{ \sqrt[3]{K_s} }  4.64

Planet X

             a_x = \sqrt[3]{ \frac{640^3}{K_s} }

             a_x = \frac{1}{ \sqrt[3]{K_s} } 74.3

Planet Y

              a_y = \sqrt[3]{ \frac{80^2}{K_s}  }

              a_y = \frac{1}{ \sqrt[3]{K_s} } 18.6

Planet z

              a_z = \sqrt[3]{ \frac{270^2}{K_s} }

              a_z = \frac{1}{ \sqrt[3]{K_s} } 41.8

From the previous results we see that planet that is farthest away is planet X

where we have used kepler's third law

3 0
3 years ago
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