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skelet666 [1.2K]
3 years ago
11

A distant galaxy emits light that has a wavelength of 434.1 nm. On earth, the wavelength of this light is measured to be 438.6 n

m. A) Decide whether this galaxy is approaching or receding from the earth. Give your reasoning. B) Find the speed of the galaxy relative to the earth.
Physics
1 answer:
maksim [4K]3 years ago
5 0

Answer:

A) receding from the earth

B) 3.078x10^6m/s

Explanation:

  • A) receding from the earth

The wavelength went from 434.1nm to 438.6nm, there was an increase in wavelength (also knowecn as redshift due to the doppler efft), this increase is due to the fact that the source that emits the radiation (the distant galaxy) is moving away and therefore the light waves it emits are "stretched", causing us to see a wavelength greater than the original.

  • B) 3.078x10^6m/s

to calculate the relative speed we use the following formula:

v_{rel}=c(1-\frac{\lambda_{1}}{\lambda_{2}} )

where c is the speed of light: c=3x10^8m/s

\lambda_{1} is the wavelength emited by the source, and

\lambda_{2} is the wavelength measured on earth.

we substitute all the values and do the calculations:

v_{rel}=(3x10^8m/s)(1-\frac{434.1nm}{438.6nm} )\\\\v_{rel}=(3x10^8m/s)(1-0.98974)\\\\v_{rel}=(3x10^8m/s)(0.01026)\\\\v_{rel}=3.078x10^6m/s

the relative speed is: 3.078x10^6m/s

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Answer:

'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.

Explanation:

The question is incomplete, find the complete question in the comment section.

Concave mirrors is an example of a curved mirror. The outer surface of a concave mirror is always coated. On the concave mirror, we have what is called the central axis or principal axis which is a line cutting through the center of the mirror. The points located on this axis are the Pole, the principal focus and the centre of curvature. <em>The focus point is close to the curved  mirror than the centre of curvature.</em>

<em></em>

During the formation of images, one of the incident rays (rays striking the plane surface) coming from the object and parallel to the principal axis, converges at the focus point after reflection because all incident rays striking the surface are meant to reflect out. <em>All incident light striking the surface all converges at a point on the central axis known as the focus.</em>

Based on the explanation above, it can be concluded that 'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.

5 0
3 years ago
What % of an object’s mass is above the water line if the object’s density is 0.82g/ml?
Sladkaya [172]

To develop this problem we will apply the Archimedes model. As well as the definitions of Weight based on mass and acceleration. The first in turn will be considered under the relationship of Density and Volume. From the values given we have to:

\rho_w =1 g/mL \rightarrow \text{Water Density}

\rho_o = 0.82g/mL \rightarrow \text{Object density}

Since it is in equilibrium, the weight of the object will have a reaction from the water, which will cause the sum of forces between the two objects to be zero, therefore

\sum F= 0

F_w-F_o = 0

F_w = F_o

m_w g = m_og

\rho_w V_w g = \rho_o V_o g

The value of gravity is canceled because it is a constant

\frac{V_w}{V_o} = \frac{\rho_o}{\rho_w}

\frac{V_w}{V_o} = \frac{0.82}{1}

\frac{V_w}{V_o} = 0.82

The portion of the object that is submerged corresponds to 82%, while the portion that is visible, above the water level will be 18%

4 0
3 years ago
How to get stud multipliers in lego star wars the skywalker saga?.
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7 0
2 years ago
A 0.9 µF capacitor is charged to a potential difference of 10.0 V. The wires connecting the capacitor to the battery are then di
jeyben [28]

Answer:

3.6μF

Explanation:

The charge on the capacitor is defined by the formula

q = CV

because the charge will be conserved

q₁ = C₁V₂

q₂ = C₂V₂ where C₂ V₂ represent the charge on the newly connected capacitor  and the voltage drop across the two capacitor will be the same

q = q₁ + q₂ = C₁V₂ + C₂V₂

CV = CV₂ + C₂V₂

CV - CV₂ = C₂V₂

C ( V - V₂) = C₂V₂

C ( V/ V₂ - V₂ /V₂) = C₂

C₂ = 0.9 ( 10 /2) - 1) = 0.9( 5 - 1) = 3.6μF

7 0
3 years ago
an object is thrown with an initial horizontal velocity of 10 meters per second and take approximately 9 seconds to reach the gr
pantera1 [17]

The horizontal velocity was constant, so:

s = vt

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s = 90

it traveled 90meters

6 0
3 years ago
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