The plate near the Artic Ridge is among those with the slower growth rates, moving slightly less than 2.5 cm/year. How much will

An object is represented by the dot on a motion map. What is the best description for the motion of this object? The object is m

ankoles [38]

The object is standing still.

5 0

3 years ago

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When the tube is filled with mercury vapor, as in this case, a sharp drop in the collected current is observed when the accelera

adell [148]

Answer:

4.9 eV .

Explanation:

It is the case of discharge through mercury tube light . In it , mercury atoms are exited due to which electrons are sent to higher energy level . Here current drops to zero because electrons are excited to higher level . Energy are absorbed in quantised manner . Energy absorbed by electrons will be 4.9 eV . That means , difference in energy between two energy level is 4.9 eV .

3 0

3 years ago

Describe a situation where you can be traveling at a low speed but have an extremely high velocity

Reil [10]

Answer:

Cruising at 35,000 feet in an airliner, straight toward the east,

at 500 miles per hour

Explanation:

3 0

3 years ago

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Light from a laser with wavelength 400 nm passed through two small openings and produces an interference pattern on a screen 1 m

alexira [117]

To solve this problem we will apply the concepts related to destructive interference from double-slit experiments. For this purpose we will define the path difference as,

$\text{Path difference}= dsin\theta = (2n-1)\frac{\lambda}{2}$

Here,

$\lambda$ = Wavelength

$\theta$ = Angle when occurs the interference point of destructive interference

Our values are given as,

$\text{Wavelength} = \lambda = 400nm = 4*10^{-7}m$

$\text{Distance of Screen} = D = 1m$

Using the previous expression we have,

$d \times \theta = \frac{\lambda}{2}$

$d \times (0.1) = \frac{4*10^{-7}}{2}$

$d = 2*10^{-6} m$

$d = 2\mu m$ $d = 2\mu m$

Therefore the distance between the two openings is $2\mu m$

8 0

3 years ago

At the very end of Wagner's series of operas The Ring of Nibelung, Brunnhilde takes the golden ring form the finger of the dead

Blababa [14]

Answer:

a) 404 m² b) apparent height = 7.5 m

Explanation:

This question is about refraction and total internal refraction.

Here I will take refractive index of air and water

$n_{air}=1\\ n_{water}=1.33=4/3$

Now let's look at the diagram I have attached here

At some angle A, the light from the ring (yellow point) under water will be totally internally refracted (B = 90°), which means that rays of light (yellow arrow) that make large enough angle A will not be able to escape from the water. Since we assumed that the ring is a point, there will be a critical cone of angle A with the ring at its apex which traces a circle of radius R on the surface of water, which, beyond this radius, no light could escape.

According to snell's law

$\frac{sin(B)}{sin(A)} = \frac{n_{water}}{n_{air}} = 4/3$

At critical angle B = 90°

$\frac{3)}{4}sin(B) = [tex]\frac{3}{4} sin(90^\circ ) = 0.75 = sin(A)$

Therefore

$A = 48.6^\circ$

With this, we can find the radius of the circle (refer to my diagram)

$h* tan (A) = R\\R =11.3 m$

And with that we can find the area

$A = \pi R^2=404\ m^2$

Additional Problem

For apparent depth from above, we can think that, since we are accustomed to seeing light at the speed of c in air, our brain interpret light from <em>any</em> source to be traveling at c. This causes light that originated under water, which has the speed of

$v_{water} = \frac{c}{n_{water}} = 0.75c$

to appear as if it has traveled with the same duration as light with speed c

In order for this to happen our brain perceive shortened length which is the apparent depth.

To put it in mathematical term

$t_{travel}=\frac{h_{apparent}}{v_{water}} =\frac{h}{c}$

So we get apparent depth

$h_{apparent}=0.75h = 7.5\ m$

4 0

3 years ago