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3241004551 [841]
3 years ago
14

How is an image produced by a plane mirror different than an image produced by a convex mirror

Physics
2 answers:
KengaRu [80]3 years ago
6 0
The light that is reflected always from you conCAVE or conVEX meaning the light that hits the mirror reflects away from you while a reflection of a plane mirror is directly reflected image without deviation
puteri [66]3 years ago
6 0

Answer:

When light impacts a plane mirror, the light ray is reflected with the same angle as the incident light ray.

For the curved mirrors, like the convex mirror, the normal surface of the mirror is not plane, so the angles change, for example, a light ray that includes horizontally in a curved part (not in the middle, a little bit above it for example) will impact a curved surface, and the reflected light ray will have a direction that "goes away" from the center of the mirror.

This means that those mirrors produce an image like a plane mirror right in the middle, and as you go to the sides, the image starts to be dilated.

You might be interested in
A quarterback passes a football from height h = 2.1 m above the field, with initial velocity v0 = 13.5 m/s at an angle θ = 32° a
SOVA2 [1]

Answer:

a)    x = v₀² sin 2θ / g

b)    t_total = 2 v₀ sin θ / g

c)    x = 16.7 m

Explanation:

This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

        sin θ = v_{oy} / vo

        cos θ = v₀ₓ / vo

         v_{oy} = v_{o} sin θ

         v₀ₓ = v₀ cos θ

         v_{oy} = 13.5 sin 32 = 7.15 m / s

         v₀ₓ = 13.5 cos 32 = 11.45 m / s

a) In the x axis there is no acceleration so the velocity is constant

         v₀ₓ = x / t

          x = v₀ₓ t

the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero

          v_{y} = v_{oy} - gt

          0 = v₀ sin θ - gt

          t = v_{o} sin θ / g

         

we substitute

       x = v₀ cos θ (2 v_{o} sin θ / g)

       x = v₀² /g      2 cos θ sin θ

       x = v₀² sin 2θ / g

at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,

b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time

at the highest point the vertical speed is zero

          v_{y} = v_{oy} - gt

          v_{y} = 0

           t = v_{oy} / g

           t = v₀ sin θ / g

as the time to get on and off is the same the total time or flight time is

           t_total = 2 t

           t_total = 2 v₀ sin θ / g

c) we calculate

          x = 13.5 2 sin (2 32) / 9.8

          x = 16.7 m

5 0
3 years ago
Can someone please help with this. "A ball is thrown vertically upward with a speed of 26.6 m/s. How high does it rise? The acce
lubasha [3.4K]

v^2=u^2+2as

0=26.6^2-2x9.8xs

-26.6^2/-2x9.8=s

calculator

8 0
3 years ago
The Huka Falls on the Waikato River is one of New Zealand's most visited natural tourist attractions. On average, the river has
OverLord2011 [107]

Answer:

(a) V = 0.75 m/s

(b) V = 0.125 m/s

Explanation:

The speed of the flow of the river can be given by following formula:

V = Q/A

V = Q/w d

where,

V = Speed of Flow of River

Q = Volume Flow Rate of River

w = width of river

d = depth of river

A = Area of Cross-Section of River = w d

(a)

Here,

Q = (300,000 L/s)(0.001 m³/1 L) = 300 m³/s

w = 20 m

d = 20 m

Therefore,

V = (300 m³/s)/(20 m)(20 m)

<u>V = 0.75 m/s</u>

<u></u>

(b)

Here,

Q = (300,000 L/s)(0.001 m³/1 L) = 300 m³/s

w = 60 m

d = 40 m

Therefore,

V = (300 m³/s)/(60 m)(40 m)

<u>V = 0.125 m/s</u>

8 0
3 years ago
Need help with ASAP please
Oksanka [162]

Answer:

first blank is chemical second blank is kinetic energy

8 0
3 years ago
If you swim with the current in a river, your speed is increased by the speed of the water; if you swim against the current, you
Blababa [14]

Answer:

11.23%

Explanation:

Lets take

Speed of man in still water =u= 1.73 m/s

Speed of flow of water = v=0.52 m/s

When swims in downward direction then speed of man = u + v

When swims in upward direction then speed of man = u - v

Lets time taken by man when he swims in downward direction is t_1 and when he swims in downward direction is t_2

Lets distance is d and it will be remain constant in both the case

d=(u+v)t_1

d=(u-v)t_2

(1.73+0.52)t_1=(1.73-0.52)t_2

t_2=1.85t_1

Time taken in still water

2 d= t x 1.73

t=1.15 x d sec

t_1=0.44d\ sec

t_2=0.82d\ sec

total time in current = 0.82 +0.44 d=1.26 d sec

So the percentage time

percentage\ time =\dfrac{1.28-1.15}{1.15}

 Percentage time =11.32%

So it will take 11.32% more time as compare to still current.

5 0
3 years ago
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