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Anon25 [30]
2 years ago
12

A narrow beam of light containing red (660 nm) and blue (470 nm) wavelengths travels from air through a 1.00-cm-thick flat piece

of crown glass and back to air again. The beam strikes at a 30.0° incident angle.
(a) At what angles do the two colors emerge?
(b) By what distance are the red and blue separated when they emerge?
Physics
1 answer:
nlexa [21]2 years ago
5 0

Answer

given,

wavelength of red light = 660 nm

wavelength of blue light = 470 nm

thickness = 1 cm = 0.01 m

angle of incident = 30°

using Snell's law

n₁ sin θ₁ = n₂ sin θ₂

refractive index for red and blue color for crown glass

n_r = 1.512     n_b = 1.524

now,

incident ray is red

sin (\theta_{ir})=\dfrac{1\times sin(30^0)}{1.512}

when incident ray is blue

sin (\theta_{ib})=\dfrac{1\times sin(30^0)}{1.524}

so,

(\theta_e)_r=sin^{-1}(\dfrac{1.512 sin (\theta_{ir})}{1})

(\theta_e)_r=sin^{-1}(\dfrac{1.512\times \dfrac{1\times sin(30^0)}{1.512}}{1})

on solving

(\theta_e)_r = 30^0

similarly for blue ray the angle of emerge is 30°

b)

now, refracting angle of blue and red ray

sin (\theta_{ir})=\dfrac{1\times sin(30^0)}{1.512}

\theta_{ir}=19.316^0

for blue ray

sin (\theta_{ib})=\dfrac{1\times sin(30^0)}{1.524}

\theta_{ib}=19.158^0

now,

 d₁ = 1 x tan(19.316°) = 0.3505 m

 d₂ = 1 x tan (19.158°) = 0.3474 m

now, the distance is separated by

  Δ d = d₁ - d₂

  Δ d = 0.3505 - 0.3474

 Δ d =0.0031 cm

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Vinil7 [7]

Answer:

d ≈ 7,6 g/cm³  

Explanation:

d = m/V = 40g/5,27cm³ ≈ 7,6 g/cm³

V = l³ = (1.74cm)³ ≈ 5,27 cm³

3 0
2 years ago
A siphon pumps water from a large reservoir to a lower tank that is initially empty. The tank also has a rounded orifice 20 ft b
trasher [3.6K]

Answer:

height of the water rise in tank is 10ft

Explanation:

Apply the bernoulli's equation between the reservoir surface (1) and siphon exit (2)

\frac{P_1}{pg} + \frac{V^2_1}{2g} + z_1= \frac{P_2}{pg} + \frac{V_2^2}{2g} +z_2

\frac{P_1}{pg} + \frac{V^2_1}{2g} +( z_1-z_2)= \frac{P_2}{pg} + \frac{V_2^2}{2g}-------(1)

substitute P_a_t_m for P_1, (P_a_t_m +pgh) for P_2

0ft/s for V₁, 20ft for (z₁ - z₂) and 32.2ft/s² for g in eqn (1)

\frac{P_1}{pg} + \frac{V^2_1}{2g} +( z_1-z_2)= \frac{P_2}{pg} + \frac{V_2^2}{2g}

\frac{P_1}{pg} + \frac{0^2_1}{2g} +( 20)= \frac{(P_a_t_m+pgh)}{pg} +\frac{V^2_2}{2\times32.2} \\\\V_2 = \sqrt{64.4(20-h)}

Applying bernoulli's equation between tank surface (3) and orifice exit (4)

\frac{P_3}{pg} + \frac{V^2_3}{2g} + z_3= \frac{P_4}{pg} + \frac{V_4^2}{2g} +z_4

substitute

P_a_t_m for P_3, P_a_t_m for P_4

0ft/s for V₃, h for z₃, 0ft for z₄, 32,2ft/s² for g

\frac{P_a_t_m}{pg} + \frac{0^2}{2g} +h=\frac{P_a_t_m}{pg} + \frac{V_4^2}{2\times32.2} +0\\\\V_4 =\sqrt{64.4h}

At equillibrium Fow rate at point 2 is equal to flow rate at point 4

Q₂ = Q₄

A₂V₂ = A₃V₃

The diameter of the orifice and the siphon are equal , hence there area should be the same

substitute A₂ for A₃

\sqrt{64.4(20-h)} for V₂

\sqrt{64.4h} for V₄

A₂V₂ = A₃V₃

A_2\sqrt{64.4(20-h)} = A_2\sqrt{64.4h}\\\\20-h=h\\\\h= 10ft

Therefore ,height of the water rise in tank is 10ft

3 0
2 years ago
A 5.00-g bullet is shot through a 1.00-kg wood block suspended on a string 2.00 m long. The center of mass of the block rises a
o-na [289]

Answer:395.6 m/s

Explanation:

Given

mass of bullet m=5 gm

mass of wood block M=1 kg

Length of string L=2 m

Center of mass rises to an height of 0.38 cm

initial velocity of bullet u=450 m/s

let v_1 and v_2 be the velocity of bullet and block after collision

Conserving momentum

mu=mv_1+Mv_2 -------------1

Now after the collision block rises to an height of 0.38 cm

Conserving Energy for block

kinetic energy of block at bottom=Gain in Potential Energy

\frac{Mv_2^2}{2}=Mgh_{cm}

v_2=\sqrt{2gh_{cm}}

v_2=\sqrt{2\times 9.8\times 0.38}

v_2=0.272 m/s

substitute the value of v_2 in equation 1

5\times 450=5\times v_1+1000\times 0.272

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2 years ago
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Answer:

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Explanation:

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The distance between the two objects, r = 2.5 x 10⁴  Km

The gravitational force between them, F = 580 N

The gravitational force between the two objects is given by the formula

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When the gravitational force becomes half, then the distance between them becomes

Let us multiply the above equation by 1/2 on both sides

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                                                   =  GMm/(√2r)²

Therefore, the distance becomes √2d, when the gravitational force between them becomes half

                                           d = √2r = √2 x 2.5 x 10⁴  Km

                                               = 3.54 x 10⁴  Km

Hence, the two objects should be kept at a distance, d = 3.54 x 10⁴  Km so that the gravitational force becomes half.

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8 0
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