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wariber [46]
3 years ago
9

g A thin uniform film of oil that can be varied in thickness covers a sheet of glass of refractive index 1.52. The refractive in

dex of the oil is 1.64. Light of wavelength 555 nm is shone from air at normal incidence on the film. Starting with no oil on the glass, you gradually increase the thickness of the oil film until the first interference maximum in the reflected light occurs. What is the thickness of the oil film at that instant
Physics
1 answer:
Makovka662 [10]3 years ago
6 0

Answer:

The right solution is "84.09 nm".

Explanation:

The given values are:

Refractive index of glass,

= 1.52

Refractive index of oil (n),

= 1.64

Wavelength (λ),

= 555 nm

Now,

The thickness of the film (t) will be:

= \frac{\lambda}{4n}

= \frac{555}{4\times 1.65}

= \frac{555}{6.6}

= 84.09 \ nm

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Answer: the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m

Explanation:

Given that;

mass of vehicle m = 1000 kg

for a low speed test; V = 2.5 m/s

bumper maximum deflection = 4 cm = 0.04 m

First we determine the energy of the vehicle just prior to impact;

W_v = 1/2mv²

we substitute

W_v = 1/2 × 1000 × (2.5)²

W_v = 3125 J

now, the the effective design stiffness k will be:

at the impact point, energy of the vehicle converts to elastic potential energy of the bumper;

hence;

W_v = 1/2kx²

we substitute

3125 = 1/2 × k (0.04)²

3125 = 0.0008k

k = 3125 / 0.0008

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Therefore, the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m

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Max hears on the news that a total solar eclipse will be visible from his town in the coming week. Which of the following explai
viktelen [127]
I believe that the answer is A but correct me if i’m wrong
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The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .
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Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is W=1.20 \cdot 10^{-3}J

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where

q is the charge

\Delta V is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

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and since the charge starts from rest, K_i = 0, so the formula becomes

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W=1.20 \cdot 10^{-3}J is the work done by the external force

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Solving the formula for \Delta V, we find

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