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3 years ago

A total solar eclipse is visible from

Physics

2 answers:

zzz [600]3 years ago

7 0

Only within the umbra of the moon's shadow.

guajiro [1.7K]3 years ago

3 0

Answer:

b. only within the moon’s umbra.

Explanation:

Solar eclipse occurs when Moon comes between the Sun and the Earth in straight line and blocks the sunlight. A shadow of moon castes on the surface of the Earth. The region passing through the moon's umbra would view a total solar eclipse. The moon completely hides the surface of the sun. It would be visible to the people residing in the bright side of the Earth within the Umbra of the moon.

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what is the wavelength λ2λ2lambda 2 of the second laser that would place its second maximum at the same location as the fourth m

nordsb [41]

The distance separating two wave crests is known as the wavelength. The wavelength of the second laser, where its second maximum would be located, is 0.109375 mm.

Double slit interference (constructive) can be described by the equation dsinθ=m λ

Here, the order is (m), the wavelength is (), and the slit distance is (d).

Information provided-

The laser's wavelength is d/8.

The slit separation is 0.500 mm in length.

Put the distance into consideration to determine the wavelength values as,

λ1=0.5 x 10⁻³/8

λ1=0.0625 x 10⁻³

Due to the second laser's wavelength, its second maximum would be at the same spot as the first laser's fourth minimum.

Therefore, we must first determine the fourth order minima as,

dsinθ= (4-1/2) λ1

dsinθ= (7/2) λ1

Using the second maxima, dsinθ= 2λ₂

Put the following value for in the equation above:

7/2λ1 = 2λ₂

7/2 x (0.0625 x 10⁻³) = 2λ₂

λ₂=0.000109375m.

Therefore, the wavelength of the second laser, where its second maximum would be located, is 0.109375 mm.

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1 year ago

The lower the value of the coefficient of friction,the_the resistance to sliding

malfutka [58]

Answer:

lower

Explanation:

The lower the value of the coefficient of friction, the lower the resistance to sliding.

The coefficient of friction is the ratio of the frictional force and the normal force pressing two surfaces in contact together.

U = $\frac{F}{N}$

U is the coefficient of friction

F is the frictional force

N is the normal force

We see that coefficient of friction is directly proportional to frictional force.

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3 years ago

jeka94

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5 0

3 years ago

3 An un calibrated mercury in glass thermometer immersed in melting ice. The length of the mercury thread is 25 mm when the ther

sammy [17]

Answer:

25 mm = 0 deg C

200 mm = 100 deg C

200 - 25 = 175 = change in thread per 100 deg C

95 - 25 = 70 mm - change in thread from 0 deg C

70 / 175 * 100 = 40 deg C final temperature at 95 mm

5 0

3 years ago

Will mark as brainliest if correct!!!!!!!

lutik1710 [3]

C. A could be ruby (speed of light = 170,000 km/s); B could be diamond (speed of light = 120,000 km/s).

Explanation:

Refraction is a phenomenon that occurs when a light rays crosses the boundary between two different mediums.

When this occurs, the light wave changes speed and also direction, according to Snell's law:

$n_1 sin \theta_1 = n_2 sin \theta_2$ (1)

where

$n_1$ is the index of refraction of the first medium

$n_2$ is the index of refraction of the second medium

$\theta_1$ is the angle of incidence (the angle between the incident ray and the normal to the boundary)

$\theta_2$ is the angle of refraction (the angle between the refracted ray and the normal to the boundary)

The index of refraction is the ratio between the speed of light in a vacuum (c) and the speed of light in the medium (v):

$n=\frac{c}{v}$

Using this definition, we can rewrite eq.(1) as

$\frac{\sin \theta_2}{\sin \theta_1} = \frac{v_2}{v_1}$

Where $v_2$ is the speed of light in the 2nd medium and $v_1$ the speed of light in the 1st medium.

Now let's analyze the situation represented in the figure: we see that as the light ray enters the 2nd medium, it bends towards the normal, this means that the angle of refraction is smaller than the angle of incidence:

$\theta_2 < \theta_1$

This means that

$\frac{\sin \theta_2}{\sin \theta_1}$

And therefore,

$\frac{v_2}{v_1}$

So, the speed of light in the second medium is smaller than the speed of light in the first medium: this occurs only in option C), which is therefore the correct choice.

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3 years ago