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3 years ago

I NEED HELP PLEASE, THANKS! :) Why does the pattern of colors repeat in a thin soap film?

Physics

1 answer:

Nadusha1986 [10]3 years ago

7 0

Answer:

When white light shines on the bubble, it acts as a prism. While some of the white light bounces back off, some of them pass through the prism forming fringes of colors. But then, they too bounce off the inner part of the film. So one set of light rays shine into a soap bubble, but two sets of rays come back out again. When they emerge, the waves that bounce off the inner film have traveled a tiny bit further than the waves that bounced off the outer film. So, we have two sets of light waves. After that, the waves starts merging (Just like the ripples in the pond) . Some add together while some cancel out. That is why we see the pattern of colors repeating in the thin film soap.

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If a machine has an efficiency of 94% and you apply 574J of work, how much work do you get out of the machine

Mariana [72]

<h3>Answer:</h3>

539.56 Joules

<h3>Explanation:</h3>

Efficiency of a machine is the ratio of work output to work input expressed as a percentage.

Efficiency = (work output/work input) × 100%
Efficiency of a machine is not 100% because so energy is lost due to friction of the moving parts and also as heat.

In this case;

Efficiency = 94%

Work input = 574 Joules

Therefore, Assuming work output is x

94% = (x/574 J) × 100%

0.94 = (x/574 J)

Thus, you get work of 539.56 J from the machine

7 0

3 years ago

A section of a parallel-plate air waveguide with a plate separation of 7.11 mm is constructed to be used at 15 GHz as an evanesc

adell [148]

Answer:

the required minimum length of the attenuator is 3.71 cm

Explanation:

Given the data in the question;

we know that;

$f_{c_1$ = c / 2a

where f is frequency, c is the speed of light in air and a is the plate separation distance.

we know that speed of light c = 3 × 10⁸ m/s = 3 × 10¹⁰ cm/s

plate separation distance a = 7.11 mm = 0.0711 cm

so we substitute

$f_{c_1$ = 3 × 10¹⁰ / 2( 0.0711 )

$f_{c_1$ = 3 × 10¹⁰ cm/s / 0.1422 cm

$f_{c_1$ = 21.1 GHz which is larger than 15 GHz { TEM mode is only propagated along the wavelength }

Now, we determine the minimum wavelength required

Each non propagating mode is attenuated by at least 100 dB at 15 GHz

Attenuation constant TE₁ and TM₁ expression is;

∝₁ = 2πf/c × √( ( $f_{c_1$ / f)² - 1 )

so we substitute

∝₁ = ((2π × 15)/3 × 10⁸ m/s) × √( (21.1 / 15)² - 1 )

∝₁ = 3.1079 × 10⁻⁷

∝₁ = 310.79 np/m

Now, To find the minimum wavelength, lets consider the design constraint;

20log₁₀ $e^{-\alpha _1l_{min$ = -100dB

we substitute

20log₁₀ $e^{-(310.7np/m)l_{min$ = -100dB

$l_{min$ = 3.71 cm

Therefore, the required minimum length of the attenuator is 3.71 cm

6 0

2 years ago

Describe a situation where your body can act as a lightning conductor.

Naddika [18.5K]

When someone is struct by lightning, the electricity passes through the body, into the earth. Here, our body acts as a lightning conductor to complete the earthing process.

7 0

2 years ago

An ice-making machine inside a refrigerator operates in a Carnot cycle. It takes heat from liquid water at 0.0 ∘C and rejects he

Agata [3.3K]

Answer:

2.36 x 10^6 J

Explanation:

Tc = 0°C = 273 K

TH = 22.5°C = 295.5 K

Qc = heat used to melt the ice

mass of ice, m = 85.7 Kg

Latent heat of fusion, L = 3.34 x 10^5 J/kg

Let Energy supplied is E which is equal to the work done

Qc = m x L = 85.7 x 3.34 x 10^5 = 286.24 x 10^5 J

Use the Carnot's equation

$\frac{Q_{H}}{Q_{c}}=\frac{T_{H}}{T_{c}}$

$Q_{H}=286.24\times 10^{5}\times \frac{295.5}{273}$

QH = 309.8 x 10^5 J

W = QH - Qc

W = (309.8 - 286.24) x 10^5

W = 23.56 x 10^5 J

W = 2.36 x 10^6 J

Thus, the energy supplied is 2.36 x 10^6 J.

8 0

3 years ago

Mass A forklift raises a box 1.2 m and does 7.0 kJ of<br> work on it. What is the mass of the box?

Mamont248 [21]

Answer:

Work and Kinetic Energy

A B

3. A 0.180 kg balls falls 2.5 m. How much work does the force of gravity do on the ball? 4.41 J

4. A forklift raises a box 1.2 m doing 7.0 kJ of work on it. What is the mass of the box? 595.24 kg

5. How much work does the force of gravity do when a 25 N object falls a distance of 3.5 m? 87.5 J

Explanation:

5 0

2 years ago