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12 days ago

Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelen

gth of d/15. The lasers produce separate interference patterns on a screen a distance 4.90 m away from the slits.
a. What is the distance Δ ymax-max between the first maxima (on the same side of the central maximum) of the two patterns?
b. What is the distance Δymax-min between the second maximum of laser 1 and the third minimum of laser 2, on the same side of the central maximum?

Physics

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You, Archimedes, suspect that the king’s crown is not solid gold but is instead gold-plated lead. To test your theory, you weigh

inna [3103]

Answer:

a) 16675.75 Kg/m³ b) 77.6%

Explanation:

The weight of the crown is 60 N, with gold's density at 19300 Kg/m³, lead's at 11340 kg/m³, water's at 1000kg/m³, and gravitational acceleration at 9.8 m/s².

The upthrust acting on the crown equals the weight in air minus its weight in water: 60 - 56.4 results in 3.6 N.

This leads to the mass of water displaced being 3.6 / 9.8, as weight equals mass times gravity.

The mass of displaced water is 0.367 Kg.

The density of water relates mass to volume as: 1000 = 0.367 / volume.

Cross-multiplying helps us determine the volume:

The crown’s volume becomes 0.367 / 1000 = 0.000367 m³ since it displaces an equal volume of water per Archimedes' principle.

Let V1 denote the gold volume and V2 the lead volume.

Total volume for the crown becomes V1 + V2.

Likewise, using the relationship of densities:

Density of gold translates to mass of gold over V1 and lead density translates to mass of lead over V2.

Thus, 19300 = mass of gold in the crown / V1 and 11340 = mass of lead in the crown / V2.

Combining them gives us: 19300 V1 = mass of gold and 11340 V2 = mass of lead.

Adding together leads to: 19300 V1 + 11340 V2 = weight of the crown / 9.8.

So, 19300 V1 + 11340 V2 = 6.12.

From V1 + V2 = 0.000367, we can express V1 in relation to V2.

V1 = 0.000367 - V2.

Substituting this into the mass equation results in:

19300 (0.000367 - V2) + 11340 V2 = 6.12.

Expanding gives:

7.083 - 19300 V2 + 11340 V2 = 6.12.

Reorganizing yields:

-7960 V2 = 6.12 - 7.083.

So, -7960 V2 = -0.963.

This leads to V2 = -0.963 / -7960 = 0.000121 (the lead volume in the crown).

Substituting V2 back into the total volume equation gives:

V1 + 0.000121 = 0.000367 m³

Thus, V1 = 0.000367 - 0.000121 = 0.000246 m³ (the gold volume in the crown).

Which leads to the mass of gold in the crown = 19300 × 0.000246 = 4.748 kg.

The mass of lead equals 11340 × 0.000121 = 1.372 kg.

The average density for the crown calculates as (mass of gold + mass of lead) / total volume = 6.12 / 0.000367 = 16675.75 kg/m³.

b) The percentage of gold by weight computes to mass of gold / total mass × 100 = approximately 77.6%.

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Mr. Chang's class participated in a science experiment where they placed three substances outside in the sunlight to determine w

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It could be B or D, but I am fairly certain it is D.

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A graph of the net force F exerted on an object as a function of x position is shown for the object of mass M as it travels a ho

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The alteration in kinetic energy is $\Delta K = 3Fd$

Clarification:

According to the work-energy principle, the task performed on an object corresponds to the alteration in its kinetic energy. In mathematical terms:

$W=K_f -K_i= \Delta K$

where:

W signifies the work performed on the object

$K_f$ denotes the kinetic energy at the end

$K_i$ indicates the kinetic energy at the start

Furthermore, when the force is exerted in line with the object’s motion, the work done is expressed as:

$W=F\Delta x$

Here,

F represents the force’s magnitude

$\Delta x$ denotes the object’s displacement

In this scenario, the force impacting the object is

While the distance moved is the horizontal length traveled, hence

$\Delta x = 3d$

Consequently, the work accomplished is

$W=(F)(3d)=3Fd$

Thus, the alteration in kinetic energy amounts to

$\Delta K = 3Fd$

Learn more about work and kinetic energy:

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Russ makes the diagram below to organize his notes about how Newton’s first law describes objects at equilibrium.

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