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Elden [556K]
3 years ago
9

The two most prominent wavelengths in the light emitted by a hydrogen discharge lamp are 656 nm(red) and 486 nm (blue). Light fr

om a hydrogen lamp illuminates a diffraction grating with 520lines/mm , and the light is observed on a screen 1.5m behind the grating.
What is the distance between the first-order red and blue fringes?
Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
STatiana [176]3 years ago
5 0

Answer:

0.152 m

Explanation:

The condition for constructive interference is

d sinθ= mλ  (m= 0,1,2,3...)

the slit width

d= 1/N

d= 10^(-3)/520

= 1.92×10^(-6)

The angular spread for the red light is

\theta_r= sin^{-1}(\frac{m\lambda}{d})

\theta_r= sin^{-1}(\frac{656\times10^{-9}}{1.92\times10^{-6}})

= 19.97°

The angular spread of blue light is

\theta_b= sin^{-1}(\frac{m\lambda}{d})

\theta_b= sin^{-1}(\frac{456\times10^{-9}}{1.92\times10^{-6}})

=14.66°

The distance between the first order red fringe and blue fringe is,

y= y_r- y_b = Ltanθ_r -Ltanθ_b

=1.50( tan19.97°-tan14.66°) = 0.152 m

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A 78.5-kg man floats in freshwater with 3.2% of his volume above water when his lungs are empty, and 4.85% of his volume above w
Dima020 [189]

Answer:

A) V_air = 1.295 L

B) Volume is not reasonable

Explanation:

A) Let;

m be total mass of the man

m_p be the mass of the man that pulled out of the water because of the buoyant force that pulled out of the lung

m_3 be the mass above the water with the empty lung

m_5 be the mass above the water with full lung

F_b be the buoyant force due to the air in the lung

V_a be the volume of air inside man's lungs

w_p be the weight that the buoyant force opposes as a result of the air.

Now, we are given;

m = 78.5 kg

m_3 = 3.2% × 78.5 = 2.512 kg

m_5 = 4.85% × 78.5 = 3.80725 kg

Now, m_p = m_5 - m_3

m_p = 3.80725 - 2.512

m_p = 1.29525 kg

From archimedes principle, we have the formula for buoyant force as;

F_b = (m_displaced water)g = (ρ_water × V_air × g)

Where ρ_water is density of water = 1000 kg/m³

Thus;

F_b = w_p = 1.29525 × 9.81

F_b = 12.7064 N

As earlier said,

F_b = (ρ_water × V_air × g)

Thus;

V_air = F_b/(ρ_water × × g)

V_air = 12.7064/(1000 × 9.81)

V_air = 1.295 × 10^(-3) m³

We want to convert to litres;

1 m³ = 1000 L

Thus;

V_air = 1.295 × 10^(-3) × 1000

V_air = 1.295 L

B) From research, the average lung capacity of an adult human being is 6 litres of air.

Thus, the calculated lung volume is not reasonable

4 0
3 years ago
Mercury's surface has many craters because _____.
Alexeev081 [22]
I think it would be that it has no atmosphere
5 0
3 years ago
A sound wave traveling downward with a speed of about 4,000 m/s suddenly slows to 1,500 m/s not far below the Earth’s surface. W
mezya [45]
<h2>Answer: an underground lake</h2>

Explanation:

In general, sound (mechanical waves) travels faster in solids than in liquids, and faster in liquids than in gases. This is because <u>the speed of the mechanical waves is determined by a relationship between the elastic properties of the medium </u>in which they are propagated and the mass per unit volume of the medium (that is:<u>density</u>).

In other words: The speed of sound varies depending on the medium through which the sound waves travel.

So, if we are told the sound wave initially had a speed of 4,000 m/s and it suddenly decreases to 1,500 m/s, this means the sound waves passed from a solid medium to a liquid medium.

Hence, the correct option is: an underground lake.

8 0
3 years ago
Read 2 more answers
Why should the safety net that is used in circuses under trapezoids be little tightened?
bixtya [17]
So that there isn't too much force restricting the structure of the circus and cause disaster
8 0
2 years ago
As a cold air mass advances on a warm air mass, what usually comes before it?​
Marina CMI [18]

Answer: A cold front occurs when a cold air mass advances into a region occupied by a warm air mass. If the boundary between the cold and warm air masses doesn't move, it is called a stationary front.

Explanation: Two types of occluded front exist: the warm-type and the cold-type. They’re distinguished by the relative temperatures of the air mass ahead of the occlusion – in other words, the air mass ahead of the original warm front – and the air mass behind the cold front. If the air behind the cold front is colder than the air ahead of the occlusion, it shoves beneath that air (because it’s denser) to form a cold-type occluded front. If the air behind the cold front is warmer than the air ahead, it rides over it to form a warm-type occluded front – which appears to be the more common case. In either situation, the lighter warm air representing the air mass originally between the warm and cold fronts sits above the boundary between the two cooler air masses.

Hope this helps!!

8 0
3 years ago
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