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Soloha48 [4]
2 years ago
13

Why do i look ugly in some mirrors but pretty in others? Which is more accurate?​

Physics
2 answers:
qwelly [4]2 years ago
6 0

Answer:   Curved mirrors often reflect a distorted image that looks different from how you look in real life. Low-quality mirrors are also likely to produce image distortions and give your face an asymmetrical appearance, which is perceived as ugly.

Explanation:

Talja [164]2 years ago
5 0

Answer:

it may be because of the angle or perception of the mirror, like for example some mirrors may be designed to warp features of the body, in my opinion don't take mirrors personally some were specifically made to trick your mind by creating warped features.

Explanation:

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What kind of energy is produce when sun reaches solar panel?
IrinaK [193]

Answer:

Radient to ElEcTrIcAAl

Explanation:

The FitnessGram Pacer Test is a multistage aerobic capacity test that progressively gets more difficult as it continues. The 20 meter pacer test will begin in 30 seconds. Line up at the start. The running speed starts slowly, but gets faster each minute after you hear this signal. A single lap should be completed each time you hear this sound. Remember to run in a straight line, and run as long as possible.

4 0
3 years ago
Read 2 more answers
A cannon shoots a ball at an angle θ above the horizontal ground. (a) Neglecting air resistance, use Newton's second law to find
nikitadnepr [17]

Answer:

a)  x = v₀ₓ t ,  y = v_{oy} t - ½ g t²

Explanation:

This is a projectile launch problem, let's use Newton's second law on each axis

X axis

       F = m a

Since there is no acceleration on the x axis, the force on this axis is zero

Y Axis  

       -W = m a_{y}

       -m g = m a_{y}  

       a_{y}  = -g

In this axis the acceleration is the acceleration of gravity

Now we can use science to find the position of the body on each axis

X axis

       x = v₀ₓ t + ½ a  t²

As the acceleration on this axis is zero

      x = v₀ₓ t

Y Axis

       y = v_{oy} t + ½ a_{y}  t²

The acceleration on this axis is –g

        y = v_{oy} t - ½ g t²

B) to find the maximum value of distance r

        r =√ x² + y²

        r = √( v₀ₓ² t² + (v_{oy} t + ½ g t²)²

We can find the maximum value of r using time respect derivatives

      dr / dt = 0

      0 = ½ 1/√( v₀ₓ² t² + (v_{oy} t + ½ g t²)²    (v₀ₓ² 2t + 2 (v_{oy} t - ½ g t²)(v_{oy} - ½ g2t)

We simplify this expression

         0 = v₀ₓ² 2t + 2 (v_{oy} t - ½ g t²)  (v_{oy} - ½ g2t)

       -v₀ₓ² t =( v_{oy} t - ½ g t²)  (v_{oy} - ½ g2t)  

      -v₀ₓ² t = v_{oy}² t -3/2 gt² v_{oy} + 1/2 g² t³  

       ½ g² t²- 3/2 g v_{oy} t  = v_{oy}² + v₀ₓ²

Let's use trigonometry to find go and vox

         sin θ = v_{oy} / v₀

         cos θ = vox / v₀

         v_{oy} = v₀ sin θ

        v₀ₓ = v₀ cos θ

We replace

         ½ g² t² -3/2 g v_{oy} t = v₀ (sin² θ +  cos² θ)

          g t² - 3v₀ sin θ t = 2 v₀/g

 

The time is maximum for the angle is zero

         

8 0
3 years ago
Margy is trying to improve her cardio endurance by performing an exercise in which she alternates walking and running 100.0 m ea
madreJ [45]
In order to answer this exercise you need to use the formulas

 S = Vo*t + (1/2)*a*t^2

Vf = Vo + at

The data will be given as

Vf = final velocity = ?

Vo = initial velocity = 1.4 m/s

a = acceleration = 0.20 m/s^2

s = displacement = 100m

And now you do the following:

100 = 1.4t + (1/2)*0.2*t^2

t = 25.388s

and

Vf = 1.4 + 0.2(25.388)

Vf = 6.5 m/s

So the answer you are looking for is 6.5 m/s
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The relationship that exists between gravity and distance and mass respectively.
ruslelena [56]

Answer:

D...............................

7 0
3 years ago
Suppose you wanted to use a non-reflecting layer for radar waves to make an aircraft invisible. What would the thickness of the
Xelga [282]

Answer:

the thickness of the film for destructive interference is 1 cm

Explanation:

We can assume that the radar wave penetrates the layer and is reflected in the inner part of it, giving rise to an interference phenomenon of the two reflected rays, we must be careful that the ray has a phase change when

* the wave passes from the air to the film with a higher refractive index

* the wavelength inside the film changes by the refractive index

         λ = λ₀ / n

so the ratio for destructive interference is

            2 n t = m λ

            t = m λ / 2n

indicate that the wavelength λ = 2 cm, suppose that the interference occurs for m = 1, therefore it is thickness

            t = 1 2/2 n

            t = 1 / n

where n is the index of refraction of the anti-reflective layer. As they tell us not to take into account the change in wavelength when penetrating the film n = 1

            t = 1 cm

So the thickness of the film for destructive interference is 1 cm

8 0
3 years ago
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