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Tanya [424]
1 year ago
9

In your own words, answer the following 3 questions about an object experiencing red shift? 1. Describe the motion of an object

experiencing red shift? 2. What is happening to its wavelength and frequency? 3. Why?
Physics
1 answer:
german1 year ago
6 0

1.  When an object is moving away from us, the light from the object is known as redshift, and when an object is moving towards us, the light from the object is known as blueshift.

2.  A wavelength increases in size, and its frequency, and energy decrease.

3.  The frequency of a wave increases, and its wavelength decreases.

Redshift is an important term for astronomers. The term can be taken literally. The wavelengths of light are stretched and perceived as shifting toward the red portion of the spectrum. The same thing happens to sound waves when the source moves relative to the observer.

As the wave frequency decreases, the wavelength increases as long as the wave velocity remains constant. If the wave speed stays the same as the frequency decreases, it means that fewer wave peaks or troughs pass through a given point in a given time period. The number of complete wavelengths in a given unit of time is called frequency. Frequency and energy decrease with increasing wavelength.

Learn more about  A wavelength here:-brainly.com/question/24452579

#SPJ1

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Answer:

The minimum stopping distance when the car is moving at

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We know by equation of motion that,

v^{2}=u^{2}+2\cdot a \cdot s

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s= Distance traveled before stop m

Case 1

u=  13 m/sec, v=0, s= 57.46 m, a=?

0^{2} = 13^{2}  + 2 \cdot a \cdot57.46

a = -1.47 m/Sec^{2} (a is negative since final velocity is less then initial velocity)

Case 2

u=29 m/sec, v=0, s= ?, a=-1.47 m/Sec^{2} (since same friction force is applied)

v^{2} = 29^{2}  - 2 \cdot 1.47 \cdot S

s = 285.94 m

Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

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