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cestrela7 [59]
2 years ago
5

A 34-cm-tall, 6.0-cm-diameter cylindrical beaker is filled to its brim with water.What is the downward force of the water on the

bottom of the beaker?
Physics
1 answer:
Solnce55 [7]2 years ago
3 0

h = height of the beaker = 34 cm = 0.34 m

ρ = density of water in the beaker = 1000 kg/m³

g = acceleration due to gravity = 9.8 m/s²

P₀ = atmospheric pressure = 101325 pa

P = pressure at the bottom of the beaker

Pressure at the bottom of the beaker is given as

P = P₀ + ρ g h

inserting the values

P = 101325 + (1000) (9.8) (0.34)

P = 104657 Pa

d = diameter of cylindrical beaker = 6 cm = 0.06 m

A = area of the bottom of the beaker = (0.25)πd² = (0.25) (3.14) (0.06)² = 0.00283 m²

Downward force of the water on the bottom of the beaker is given as

F = PA

F = (104657) (0.00283)

F = 296.2 N

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option ( D )correct

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The refractive index n of transparent acrylic plastic (full name Poly(methyl methacrylate)) depends on the color (wavelength) of
Novosadov [1.4K]

Answer:

The angle between the blue beam and the red beam in the acrylic block is  

 \theta _d  =0.19 ^o

Explanation:

From the question we are told that

     The  refractive index of the transparent acrylic plastic for blue light is  n_F  =  1.497

     The  wavelength of the blue light is F  =  486.1 nm  =  486.1 *10^{-9} \ m

    The  refractive index of the transparent acrylic plastic for red light is  n_C  =  1.488

       The  wavelength of the red light is C =  656.3 nm  = 656.3 *10^{-9} \  m

    The incidence angle is  i  =  45^o

Generally from Snell's law the angle of refraction of the blue light  in the acrylic block  is mathematically represented as

       r_F =  sin ^{-1}[\frac{sin(i) *  n_a }{n_F} ]

Where  n_a is the refractive index of air which have a value ofn_a =  1

So

     r_F =  sin ^{-1}[\frac{sin(45) *  1 }{ 1.497} ]

      r_F  =  28.18^o

Generally from Snell's law the angle of refraction of the red light in the acrylic block is mathematically represented as

       r_C =  sin ^{-1}[\frac{sin(i) *  n_a }{n_C} ]

Where  n_a is the refractive index of air which have a value ofn_a =  1

So

     r_C =  sin ^{-1}[\frac{sin(45) *  1 }{ 1.488} ]

      r_F  =  28.37^o

The angle between the blue beam and the red beam in the acrylic block

     \theta _d  =  r_C  - r_F

substituting values

       \theta _d  = 28.37 -  28.18

       \theta _d  =0.19 ^o

 

4 0
2 years ago
A ray of light is incident on a plane surface separating two sheets of glass with refractive indexes 1.70 and 1.58. The angel of
dlinn [17]

Answer:

r = 71.8⁰

Explanation:

given,

refractive index of the glass 1 = 1.70

refractive index of glass 2 = 1.58

angle of incidence = 62°

angle of refraction =?

using Snell's law

\dfrac{sin\ i}{sin\ r} = \dfrac{n_2}{n_1}

\dfrac{sin\ 62^0}{sin\ r} = \dfrac{1.58}{1.70}

1.7 ×sin 62 ^0 = 1.58× sin r

sinr = \dfrac{1.7\times sin 62^0}{1.58}

sin r = 0.95

r = sin⁻¹(0.95)

r = 71.8⁰

angle of refraction =r = 71.8⁰

4 0
2 years ago
Read 2 more answers
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