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UNO [17]
2 years ago
7

Why might the term red hot be misleading? (relating to stars)

Physics
1 answer:
Allushta [10]2 years ago
3 0

red hot is hot, but other colours are even hotter. stars may be hotter than red hot.

there is also something calle the red shift.

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when white light is incident on prism, which one of the resulting color components will have the lowest index of refraction?
attashe74 [19]
The color components that will have the lowest index of refraction will be orange.
3 0
3 years ago
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Forces applied in
Anni [7]

Answer:

A -Added when in the same direction

    Subtracted when in opposite directions.

Explanation:

8 0
2 years ago
An element has the following natural abundances and isotopic masses: 90.92% abundance with 19.99 amu, 0.26% abundance with 20.99
sashaice [31]

<u>Answer:</u> The average atomic mass of the given element is 20.169 amu.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of the isotopes each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i     .....(1)

We are given:

  • For isotope 1:

Mass of isotope 1 = 19.99 amu

Percentage abundance of isotope 1 = 90.92 %

Fractional abundance of isotope 1 = 0.9092

  • For isotope 2:

Mass of isotope 2 = 20.99 amu

Percentage abundance of isotope 2 = 0.26%

Fractional abundance of isotope 2 = 0.0026

  • For isotope 3:

Mass of isotope 3 = 21.99 amu

Percentage abundance of isotope 3 = 8.82%

Fractional abundance of isotope 3 = 0.0882  

Putting values in equation 1, we get:

\text{Average atomic mass}=[(19.99\times 0.9092)+(20.99\times 0.0026)+(21.99\times 0.0882)]

\text{Average atomic mass}=20.169amu

Hence, the average atomic mass of the given element is 20.169 amu.

4 0
3 years ago
Why do you keep moving forward when you slam on the brakes of your bike?
g100num [7]
Because the act of braking is an example of negative acceleration.
Example: if the rate of braking was say 2 meters per second^2, and the starting velocity was 10 m/s, it would take 5 seconds to come to a stop(during those 5 seconds you would still be moving).
6 0
3 years ago
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An object is placed in front of a diverging lens, such that the object-to-image distance is 71 cm.
Pachacha [2.7K]

Explanation:

Given that,

Object-to-image distance d= 71 cm

Image distance = 26 cm

We need to calculate the object distance

u -v= d

u=71+26=97\ cm

We need to calculate the focal length

Using formula of lens

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}

put the value into the formula

\dfrac{1}{f}=\dfrac{1}{-26}+\dfrac{1}{97}

\dfrac{1}{f}=-\dfrac{71}{2522}

f=-35.52\ cm

The focal length of the lens is 35.52.

(B). Given that,

Object distance = 95 cm

Focal length = 29 cm

We need to calculate the distance of the image

Using formula of lens

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}

Put the value in to the formula

\dfrac{1}{-29}=\dfrac{1}{v}-\dfrac{1}{95}

\dfrac{1}{v}=\dfrac{1}{-29}-\dfrac{1}{95}

\dfrac{1}{v}=-\dfrac{124}{2755}

v=-22.21\ cm

We need to calculate the magnification

Using formula of magnification

m=\dfrac{v}{u}

m=\dfrac{22.21}{95}

m=0.233

The magnification is 0.233.

The image is virtual.

Hence, This is the required solution.

4 0
3 years ago
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