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GREYUIT [131]
3 years ago
15

A ray of laser light strikes a glass surface at an angle a=25.0° tot he normal and it is refracted to an angle b=18.5° to the no

rmal. What is the index of refraction nb for this type of glass? Na=1.00
A. 3.26
B. 3.52
C. 1.32
D. 1.89
Physics
1 answer:
USPshnik [31]3 years ago
3 0

Answer:

1.33

Explanation:

According to snell's law,

n = sin(a)/sin(b)

Given

a = 25.0°

b = 18.5°

Substitute

n = sin25/sin18.5

n = 0.4226/0.3173

n = 1.33

Hence the index of refraction nb for this type of glass is 1.33

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The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 23
Burka [1]

Answer:

a) v(2) = 3.9\,\frac{m}{s}, b) v(4) = -15.7\,\frac{m}{s}

Explanation:

a) The equation for vertical velocity is obtained by deriving the function with respect to time:

v(t) = 23.5 -9.8\cdot t

The velocities at given instants are, respectivelly:

v(2) = 3.9\,\frac{m}{s}

v(4) = -15.7\,\frac{m}{s}

8 0
3 years ago
The amount of the lighted side of the moon you can see is the same during
attashe74 [19]

<u>Answer:</u>

The amount of the lighted side of the moon you can see is the same during "how much of the sunlit side of the moon faces Earth".

<u>Explanation:</u>

The Moon is in sequential rotation with Earth, and thus displays the Sun, the close side, always on the same side. Thanks to libration, Earth can display slightly greater than half (nearly 59 per cent) of the entire lunar surface.

The side of the Moon facing Earth is considered the near side, and the far side is called the reverse. The far side is often referred to as the "dark side" inaccurately but it is actually highlighted as often as the near side: once every 29.5 Earth days. During the New Moon the near side becomes blurred.

5 0
3 years ago
Ans of this question A test charge of 1 couloumb moved from 30cm against the field of intensity 50N/c find the energy store in i
UkoKoshka [18]

Answer:

A. Zero

Explanation:

Given data,

The charge of the test charge, q = 1 C

The distance the charge moved against the filed of intensity, x = 30 cm

                                                                                                        = 0.3 m

The electric field intensity, E = 50 N/C

The energy stored in the charge at 0.3 m is given by the formula,

                                V = k q/r

Where,                        

                                     = 9 x 10⁹ Nm²C⁻²

The charge is moved from the potential V₁ to V₂ at 30 cm

Substituting the given values in the above equation

                            V₁ = 9 x 10⁹ x 30 / 0.3

                                =  1.5 x 10¹² J

And,

                            V₂ = 1.5 x 10¹² J

The energy stored in it is,

                             W = V₂ - V₁

                                  = 0

Hence, the energy stored in the charge is, W = 0        

6 0
3 years ago
A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of 152 om and makes an angl
aliya0001 [1]

Answer:

D₂= 167,21 cm : Magnitude  of the second displacement

β= 21.8° , countercockwise from the positive x-axis: Direction of the second displacement

Explanation:

We find the x-y components for the given vectors:

i:  unit vector in x direction

j:unit vector in y direction

D₁: Displacement Vector 1

D₂: Displacement Vector 2

R= resulta displacement vector

D₁= 152*cos110°(i)+152*sin110°(j)=-51.99i+142.83j

D₂= -D₂(i)-D₂(j)

R=  131*cos38°(i)+ 131*sin38°(j) = 103.23i+80.65j

We propose the vector equation for sum of vectors:

D₁+ D₂= R

-51.99i+142.83j+D₂x(i)-D₂y(j) = 103.23i+80.65j

-51.99i+D₂x(i)=103.23i

D₂x=103.23+51.99=155.22 cm

+142.83j-D₂y(j) =+80.65j

D₂y=142.83-80.65=62.18 cm

Magnitude and direction of the second displacement

D_{2} =\sqrt{(D_{x})^{2} +(D_{y} )^{2}  }

D_{2} =\sqrt{(155.22)^{2} +(62.18 )^{2}  }

D₂= 167.21 cm

Direction of the second displacement

\beta = tan^{-1} \frac{D_{y}}{D_{x} }

\beta = tan^{-1} \frac{62.18}{155.22 }

β= 21.8°

D₂= 167,21 cm : Magnitude  of the second displacement

β= 21.8.° , countercockwise from the positive x-axis: Direction of the second displacement

6 0
3 years ago
An object was thrown at a certain angle above the ground.It reaches a maximum height of 42.50 meters and hits back the ground 76
aksik [14]

Answer:

Explanation:  Please see my attached calculations.

7 0
3 years ago
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