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s2008m [1.1K]
3 years ago
10

What is the maximum number of lines per centimeter a diffraction grating can have and produce a complete first-order spectrum fo

r visible light? Assume visible light has a range of 400 nm to 700 nm.
Physics
1 answer:
Pachacha [2.7K]3 years ago
5 0

Answer:

14,300 lines per cm

Explanation:

Answer:

14,300 cm per line

Explanation:

λ400 nm to 400nm

We can find the maximum number of lines per centimeter, which is reciprocal of the least distance separating two adjacent slits, using the following equation.

mλ = dsin (θ)

In this equation,

m is the order of diffraction.

λ is the wavelength of the incident light.

d is the distance separating the centers of the two slits.

θ is the angle at which the mth order would diffract.

To find the least separation that allows the observation of one complete order of spectrum of the visible region, we use the maximum wavelength of the visible region is 700 nm.

d =  mλ / sin (θ)

As we want the distance d to be the smallest then sin (θ) must be the greatest, and the greatest value of the sin (θ) is 1. For that we also use the longest wavelength because using the smallest wavelength, the longest wavelength would not be diffracted.

d =  mλ / sin (θ)

d =  1 x 700nm / 1

  = 700 nm

So, the least separation that would allow for the possibility of observing complete first order of the visible region spectra is 700 nm, and knowing the least separation we can find the maximum number of lines per cm, which is the reciprocal of the number of lines per cm.

n = 1/d

   = 1 / 700 x 10^{-9}

  = 1, 430,000 lines per m  

  =  14,300 lines per cm

<u>The maximum number of lines per cm, that would allow for the observation of the complete first order visible spectra.</u>

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pantera1 [17]

<u>The following statements are false about collisions: </u>

  • The velocity change of two respective objects involved in a collision will always be equal.
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Answer: Option B, and D

<u>Explanation: </u>

In any collisions, equal amount of net force will be acted upon the colliding objects due to the third law of Newton, irrespective of the significance difference in mass of the objects. Similarly, they can also have different acceleration values during collision of two objects if the masses are identical.

But the statements regarding the equal change in velocity of two objects respectively involved in collision always is false, as the conservation of momentum is applicable for isolated system only. So it is true for only isolated system and not in all the systems.

The same reason goes for falsifying the fourth statement which states that total momentum is always conserved between two objects involved in a collision as this statement is only true for isolated system where the conservation of momentum can be applied. Thus the second and fourth statement is false regarding collision.

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3 years ago
Pink
FromTheMoon [43]

Answer:

Pink - Pluto

Green - Mercury

Red - Mars

Black - Saturn

Blue - Neptune

Yellow - Jupiter

Orange - Jupiter

Milk Color - Venus

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4 years ago
A 50 g copper calorimeter contains 250 g of water at 20 C. How much steam be condensed into the water to make the final temperat
Nostrana [21]

Answer:

Approximately 13\; \rm g of steam at 100\; \rm ^\circ C (assuming that the boiling point of water in this experiment is 100\; \rm ^\circ C\!.)

Explanation:

Latent heat of condensation/evaporation of water: 2260\; \rm J \cdot g^{-1}.

Both mass values in this question are given in grams. Hence, convert the specific heat values from this question to \rm J \cdot g^{-1}.

Specific heat of water: 4.2\; \rm J \cdot g^{-1}\cdot \rm K^{-1}.

Specific heat of copper: 0.39\; \rm J \cdot g^{-1}\cdot K^{-1}.

The temperature of this calorimeter and the 250\; \rm g of water that it initially contains increased from 20\; \rm ^\circ C to 50\; \rm ^\circ C. Calculate the amount of energy that would be absorbed:

\begin{aligned}& Q(\text{copper}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 0.39\; \rm J \cdot g^{-1}\cdot K^{-1} \times 50\; \rm g \times (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 585\; \rm J  \end{aligned}.

\begin{aligned}& Q(\text{cool water}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; \rm J \cdot g^{-1}\cdot K^{-1} \times 250\; \rm g \times (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 31500\; \rm J  \end{aligned}.

Hence, it would take an extra 585\; \rm J + 31500\; \rm J = 32085\; \rm J of energy to increase the temperature of the calorimeter and the 250\; \rm g of water that it initially contains from 20\; \rm ^\circ C to 50\; \rm ^\circ C.

Assume that it would take x grams of steam at 100\; \rm ^\circ C ensure that the equilibrium temperature of the system is 50\; \rm ^\circ C.

In other words, x\; \rm g of steam at 100\; \rm ^\circ C would need to release 32085\; \rm J as it condenses (releases latent heat) and cools down to 50\; \rm ^\circ C.

Latent heat of condensation from x\; \rm g of steam: 2260\; {\rm J \cdot g^{-1}} \times (x\; {\rm g}) = (2260\, x)\; \rm J.

Energy released when that x\; {\rm g} of water from the steam cools down from 100\; \rm ^\circ C to 50\; \rm ^\circ C:

\begin{aligned}Q = \;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; {\rm J \cdot g^{-1}\cdot K^{-1}} \times (x\; \rm g) \times (100\;{\rm ^\circ C} - 50\;{\rm ^\circ C}) \\ =\; & (210\, x)\; \rm J  \end{aligned}.

These two parts of energy should add up to 32085\; \rm J. That would be exactly what it would take to raise the temperature of the calorimeter and the water that it initially contains from 20\; \rm ^\circ C to 50\; \rm ^\circ C.

(2260\, x)\; {\rm J} + (210\, x)\; {\rm J} = 32085\; \rm J.

Solve for x:

x \approx 13.

Hence, it would take approximately 13\; \rm g of steam at 100\; \rm ^\circ C for the equilibrium temperature of the system to be 50\; \rm ^\circ C.

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Eduardwww [97]

Complete Question

A proton is located at <3 x 10^{-10}, -5*10^{-10}  , -5*10^{-10}> m. What is r, the vector from the origin to the location of the proton

Answer:

The  vector position is   \=r=

Explanation:

From the question we are told that

  The  position of the proton is m

Generally the vector location of the proton is mathematically represented as

       \= r =

So substituting values

     \=r=

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frez [133]

Answer:

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Explanation:

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