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sweet-ann [11.9K]
3 years ago
5

What is the critical angle of light traveling from benzene (n=1.501) into air?

Physics
1 answer:
Taya2010 [7]3 years ago
4 0
The critical angle formula should be: sin^-1(1/n) 

 where "n" is 1.501 into the air

<span>The critical angle of light travelling from benzene, happens because the larger angles of incidence from the inside of the benzene has experienced the total internal reflection. </span>


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A box slides down a frictionless ramp.if it starts at rest, what is it’s speed at the bottom?
zhenek [66]

8.854 m/s is the speed of the box after it reaches bottom of the ramp.

<u>Explanation</u>:

From the figure we came to know that height of the block is 4 m.

We know that,

Total "initial energy of an object" = Total "final energy of an object "

Total "initial energy of an object" is = "sum of potential energy" and "kinetic energy" of an object at its initial position.

\text { "g" acceleration due to gravity is } 9.8 \mathrm{m} / \mathrm{s}^{2}

\text { Total initial energy }=\mathrm{m} \times \mathrm{g} \times \mathrm{h}_{\mathrm{i}}+\frac{1}{2} \mathrm{m} v_{i}^{2}

Initial velocity is “0” as the object does not have starting speed

\text { Height of the block where the object is placed initially }\left(h_{i}\right) \text { is } 4 \mathrm{m} \text { . }

\text { Total initial energy }=\mathrm{m} \times 9.8 \times 4+\frac{1}{2} \mathrm{m} 0^{2}

Total initial energy = 39.2 × m

\text { Total final energy }=\mathrm{m} \times \mathrm{g} \times \mathrm{h}_{\mathrm{f}}+\frac{1}{2} m v_{f}^{2}

\text { We need to find final velocity } v_f

\text { Height of the block where the object is travelled to bottom (h_) is } 0 \mathrm{m} \text { . }

\text { Total final energy }=\mathrm{m} \times 9.8 \times 0+\frac{1}{2} m v_{f}^{2}

Now,  Total initial energy of an object = Total final energy of an object

39.2 \times \mathrm{m}=0.5 \mathrm{m} v_{f}^{2}

\frac{39.2}{0.5}=v_{f}^{2}

v_{f}^{2}=78.4

v_{f}=\sqrt{78.4}

v_{f}=8.854 \mathrm{m} / \mathrm{s}

Final speed is 8.854 m/s.

3 0
3 years ago
2) A ray of light in air is approaching the boundary with water at an
KiRa [710]

Answer:

Explanation:

ASSUMING the 52° is the angle of incidence measured from the perpendicular to the surface

          n₁sinθ₁ = n₂sinθ₂

          1 sin52 = 1.33sinθ₂

                  θ₂ = arcsin(sin52 / 1.33)

                  θ₂ = 36°

as measured from the perpendicular to the surface

5 0
2 years ago
Hey can someone send me the answer to this
bagirrra123 [75]

Cars 'A' and 'C' look like they're moving at the same speed.  If their tracks are parallel, then they're also moving with the same velocity.

5 0
3 years ago
The potential energy of a pair of hydrogen atoms separated by a large distance x is given by U(x)=−C6/x6, where C6 is a positive
kirza4 [7]

Answer:

The correct answer will be "-\frac{6C_{6}}{x^{7}}". The further explanation is given below.

Explanation:

The potential energy will be,

⇒  U(x)= -\frac{C_{6}}{x^6}

The expression of force will be,

⇒  F=-\frac{dU(x)}{dx}

⇒      =-(C_{6}(-6)x^{-7})

⇒      =-\frac{6C_{6}}{x^{7}}

Force seems to be appealing because the expression has been negative. It therefore means that the force or substance is acting laterally in on itself.

6 0
3 years ago
For crystal diffraction experiments, wavelengths on the order of 0.25 nm are often appropriate.
Kamila [148]

Answer:

A) E = 4.96 x 10³ eV

B) E = 4.19 x 10⁴ eV

C) E = 3.73 x 10⁹ eV

Explanation:

A)

For photon energy is given as:

E = hv

E = \frac{hc}{\lambda}

where,

E = energy of photon = ?

h = 6.625 x 10⁻³⁴ J.s

λ = wavelength = 0.25 nm = 0.25 x 10⁻⁹ m

Therefore,

E = \frac{(6.625 x 10^{-34} J.s)(3 x 10^8 m/s)}{0.25 x 10^{-9} m}

E = (7.95 x 10^{-16} J)(\frac{1 eV}{1.6 x 10^{-19} J})

<u>E = 4.96 x 10³ eV</u>

<u></u>

B)

The energy of a particle at rest is given as:

E = m_{0}c^2

where,

E = Energy of electron = ?

m₀ = rest mass of electron = 9.1 x 10⁻³¹ kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

E = (9.1 x 10^{-31} kg)(3  x  10^8 m/s)^2\\

E = (8.19 x 10^{-14} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\

<u>E = 4.19 x 10⁴ eV</u>

<u></u>

C)

The energy of a particle at rest is given as:

E = m_{0}c^2

where,

E = Energy of alpha particle = ?

m₀ = rest mass of alpha particle = 6.64 x 10⁻²⁷ kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

E = (6.64 x 10^{-27} kg)(3  x  10^8 m/s)^2\\

E = (5.97 x 10^{-10} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\

<u>E = 3.73 x 10⁹ eV</u>

8 0
2 years ago
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