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ZanzabumX [31]
3 years ago
6

The red light from the ant passed into the lens

Physics
1 answer:
s344n2d4d5 [400]3 years ago
6 0

<u>Answer:</u>

<em>When red light enters the lens its wavelength decreases but its frequency remains the same. </em>

<u>Explanation:</u>

Lens is made of glass which is a <em>denser medium</em> when compared with air. Thus light will undergo refraction and bend toward the normal. On entering  a denser medium from rarer medium the speed of light decreases.

<em>Speed, frequency and wavelength of an electromagnetic wave</em> are related according to the equation.

<em>                                                                 v=ϑλ </em>

Where is <em>v is the frequency,  λ is the wavelength and v is the speed of the em wave. </em>

On entering the lens the wavelength of the red light as well as speed decreases but the<em> frequency remains the same. </em>

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Heptane is always composed of 84.0% carbon and 16.0% hydrogen. This illustrates the law of multiple proportions. conservation of
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Heptane is always composed of 84.0% carbon and 16.0% hydrogen. This illustrates the "law of definite proportions".

Answer: Option C

<u>Explanation:</u>

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5 0
3 years ago
Two blocks with different temperatures had entropies of 10 J/K and 30 J/K before they were brought in contact. What can you say
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Answer:

a. Ssystem  > 40 J/K

Explanation:

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The entropy of first block = 10 J/K

The entropy of second block = 30 J/K

When two bodies come into contact with each other, the entropy of the combined system will increase and the entropy sum remains unchanged: According to the Second law of thermodynamics.The entropy of the system will be greater than 40  J/K.

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6 0
3 years ago
A 49.0 kg wheel, essentially a thin hoop with radius 0.730 m, is rotating at 114 rev/min. It must be brought to a stop in 22.0 s
belka [17]

Explanation:

Mass of the wheel, m = 49 kg

Radius of the hoop, r = 0.73 m

Initial angular speed of the wheel, \omega_i=114\ rev/min = 11.93\ rad/s

Final angular speed of the wheel, \omega_f=0

Time, t = 22 s

(a) If I is the moment of inertia of the hoop. It is equal to,

I=mr^2

I=49\times (0.73)^2

I=26.11\ kg-m^2

We know that the work done is equal to change in kinetic energy.

W=\Delta E

W=\dfrac{1}{2}I(\omega_f^2-\omega_i^2)

W=-\dfrac{1}{2}\times 26.11\times (11.93^2)

W = -1858.05 Joules

(b) Let P is the average power. It is given by :

P=\dfrac{W}{t}

P=\dfrac{1858.05\ J}{22\ s}

P =84.45 watts

Hence, this is the required solution.

4 0
3 years ago
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