1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Alona [7]
3 years ago
8

The top of a swimming pool is at ground level. If the pool is 3.00 m deep, how far below ground level does the bottom of the poo

l appear to be located for the following conditions? (The index of refraction of water is 1.333.)
(a) The pool is completely filled with water.
______m below ground level

(b) The pool is filled halfway with water.
______m below ground level
Physics
1 answer:
Yuri [45]3 years ago
4 0

Answer:

a) 2.25 m

b) 2.625 m

Explanation:

Refraction is the name given to the phenomenon of the speed of light changing the the boundary when it moves from one physical medium to the other.

Refractive index is the ratio of the speed of light in empty vacuum (air is an appropriate substitution) to the speed of light in the medium under consideration.

In terms of real and apparent depth, the refractive index is given by

η = (real depth)/(apparent depth)

a) Real depth = 3.00 m

Apparent depth = ?

Refractive index, η = 1.333

1.333 = 3/(apparent depth)

Apparent depth = 3/1.3333 = 2.25 m.

Hence the bottom of the pool appears to be 2.25 m below the ground level.

b) Real depth = 1.5 m

Apparent depth = ?

Refractive index, η = 1.333

1.3333 = 1.5/(apparent depth)

Apparent depth = 1.5/1.3333 = 1.125 m

But the pool is half filled with water, there is a 1.5 m depth on top of the pool before refraction starts.

So, apparent depth of the pool = 1.5 + 1.125 = 2.625 m below the ground level

You might be interested in
A straight line with a positive slope on a velocity-time graph indicates which of the following?
V125BC [204]

Answer:

constant acceleration

Explanation:

I may be wrong, but I'm almost positive cuz I am also taking physics. ;)

4 0
3 years ago
A 37.5 kg box initially at rest is pushed 4.05 m along a rough, horizontal floor with a constant applied horizontal force of 150
vodomira [7]

Answer:

a) 607.5 J

b) 160.531875 J

c)  0 J

d)  0 J

e) 2.925 m\s

Explanation:

The given data :-

  • Mass of the box ( m ) = 37.5 kg.
  • Displacement made by box ( x ) = 4.05 m.
  • Horizontal force ( F ) = 150 N.
  • The co-efficient of friction between box and floor ( μ ) = 0.3
  • Gravitational force ( N ) = m × g = 37.5 × 9.81 = 367.875

Solution:-

a) The work done by applied force ( W )

W = force applied × displacement = 150 × 4.05 = 607.5 J

b)  The increase in internal energy in the box-floor system due to friction.

Frictional force ( f ) = μ × N = 0.3 × 367.875 = 110.3625 N

Change in internal energy = change in kinetic energy.

ΔU = ( K.E )₂ - ( K.E )₁

Since the initial velocity is zero so the  ( K.E )₁ = 0  

ΔU = ( K.E )₂ = ( F - f ) × ( x ) = ( 150 - 110.3625 ) × 4.05 = 160.531875 J

c) The work done by the normal force .

Displacement of box vertically = 0

W = force applied × displacement = 367.875 × 0 = 0 J

d)  The work done by the gravitational force.

Displacement of box vertically = 0

W = force applied × displacement = 367.875 × 0 = 0 J

e) The change in kinetic energy of the box

( K.E )₂ - ( K.E )₁ = ( K.E )₂ - 0 = ( F - f ) × ( x ) = ( 150 - 110.3625 ) × 4.05 = 160.531875 J

f) The final speed of the box

( K.E )₂ = 160.531875 J = 0.5 × 37.5 × v²

v² = 8.56

v = 2.925 m\s.

5 0
3 years ago
Find the volume of a box measuring 2cm by 7 cm by 3cm
PolarNik [594]

Answer:

42cm

Explanation:

To find the volume of a box, you simply multiply the length, by the width, then by the heigh!

2cm x 7cm x 3cm = 42cm

I hope my answer helped you! :)

4 0
3 years ago
Read 2 more answers
Two large metal plates of area 1.2 m^2 face each other, 6.5 cm apart, with equal charge magnitudes but opposite signs. The field
andrey2020 [161]

Answer:

Q= 722.5 *10⁻¹² C

Explanation:

Conceptual analysis

For a parallel plate capacitor, we can use the following formula :

E= (Q) /(ϵ₀*A)  Formula (1)

Where:

E: electric field between the plates ( N/C)

Q: Charge of  the plates (C)

ϵo : vacuum permittivity  ( C²/ N.m²)

A :  area oh the plates (m²)

Known data

A = 1.2 m²

E= 68 N/C

ϵo= 8.8542*10⁻¹² (  C²/ N.m²)

Problem development

We apply the formula (1) :

68= \frac{Q}{(8.8542*10^{-12} )(1.2)}

Q= (68) (8.8542*10⁻¹²)(1.2)

Q= 722.5 *10⁻¹² C

6 0
3 years ago
A 30.0 kg object rests on a flat, frictionless surface. A rope lifts up on the object with a force of 309 N. What is the acceler
saveliy_v [14]
Total force = Frope - mg = 309-300
total force = ma
9 = 30a
4 0
3 years ago
Other questions:
  • a defensive tackle picks up the 0.5kg football to a height of 0.8m in 0.25s .... 1.calculate the work done 2.calculate the power
    7·1 answer
  • An object that is in free fall seems to be
    12·1 answer
  • Which of the following ways is usable energy lost?
    14·2 answers
  • A ___ is one way a star might die and it can trigger the beginning of a new stars life cycle .
    11·1 answer
  • The orbital velocity of the Earth about the sun is 30 km/s. If the Earth were suddenly stopped in its tracks, it would simply fa
    9·1 answer
  • Suppose you are in a spaceship traveling at 99% of the speed of light past a long, narrow space station. Your direction of trave
    8·2 answers
  • Name all nfl players
    14·1 answer
  • One of the games at a carnival involves trying to ring a bell with a ball by hitting a lever that propels the ball into the air.
    14·1 answer
  • A dog ran 10 meters in 2 seconds. What was the dog's speed? (Use the
    13·2 answers
  • What property of matter changes depending on location?
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!