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3 years ago

A mirage is created when light is refracted ___.

Physics

2 answers:

omeli [17]3 years ago

7 0

Answer:

through layers of hot air just above a surface,causing it to follow a curved path.

elena-14-01-66 [18.8K]3 years ago

5 0

. . . downward (through the air).

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How far does a car travel as it accelerates from rest to 24 m/s in 6 seconds?

____ [38]

Answer:

$\boxed{\sf Distance \ travelled = 72 \ m}$

Given:

Initial velocity (u) = 0 m/s (Starts from rest)

Final velocity (v) = 24 m/s

Time taken (t) = 6 seconds

To Find:

Distance travelled by car (s)

Explanation:

From equation of motion of object moving with uniform acceleration in straight line we have:

$\boxed{ \bold{s = (\frac{v + u}{2} )t}}$

By substituting value of v, u & t in the equation we get:

$\sf \implies s = ( \frac{24 + 0}{2} ) \times 6 \\ \\ \sf \implies s = \frac{24}{2} \times 6 \\ \\ \sf \implies s = 12 \times 6 \\ \\ \sf \implies s = 72 \: m$

$\therefore$

Distance travelled by car (s) = 72 m

7 0

4 years ago

What is the purpose of control group ?

kirill [66]

Control group in a scientific experiment is a group separated from the rest of the experiment, it is where the independent variable being tested cannot influence the results. It isolates the independent variable's effects on the experiment and can helps rule out alternative explanations of the experimental results.

6 0

3 years ago

Q: A: Sarah and Maria made a telephone using two cans and a string. It was a very long string. Maria went upstairs in her house

GrogVix [38]

This contraption is a lot of fun, and you really should try it
some time.

The sound waves move from one can to the other one
by traveling through the string.

3 0

3 years ago

Read 2 more answers

The electric field at the distance of 3.5 meters from an infinite wall of charges is 125 N/C. What is the magnitude of the elect

RideAnS [48]

Explanation:

It is given that,

Distance, r = 3.5 m

Electric field due to an infinite wall of charges, E = 125 N/C

We need to find the electric field 1.5 meters from the wall, r' = 1.5 m. Let it is equal to E'. For an infinite wall of charge the electric field is given by :

$E=\dfrac{\lambda}{2\pi \epsilon_o r}$

It is clear that the electric field is inversely proportional to the distance. So,

$\dfrac{E}{E'}=\dfrac{r'}{r}$

$E'=\dfrac{Er}{r'}$

$E'=\dfrac{125\times 3.5}{1.5}$

E' = 291.67 N/C

So, the magnitude of the electric field 1.5 meters from the wall is 291.67 N/C. Hence, this is the required solution.

5 0

4 years ago

A point source of light is 1.24 m below the surface of a pool. What is the diameter of the circle of light that a person above t

Ksivusya [100]

Answer:

D = 2.828 m

Explanation:

given,

distance of source of light = 1.24 m below surface of pool

refractive index of the water = n₁ = 1.33

refractive index of air = n₂ = 1

refraction angle be = 90°

let C be the critical angle

Radius = d tan C

d is the depth of the source

Using Snell's law

n₁ sin C = n₂ sin R

1.33 x sin C = 1 x sin 90°

$sin C = \dfrac{1}{1.33}$

$C = sin^{-1}(0.752)$

C = 48.75°

hence,

R = 1.24 x tan 48.75°

R = 1.414 m

Diameter = 2 x R

D = 2 x 1.414

D = 2.828 m

5 0

3 years ago