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Lelechka [254]
3 years ago
6

A star has a spectrum shift exhibiting a red shift. What can you conclude about the distance from earth to that star

Physics
2 answers:
Akimi4 [234]3 years ago
6 0
Red shift in the spectrum of a star is always taken to mean that the source of the light is moving away from us.

That's the whole basis for the notion of the expanding universe, since we don't know of any other mechanism that can shift light toward longer wavelengths.
sergejj [24]3 years ago
5 0
Its going away cuz red shift goes away (somehow i think of hell so think that u wanna go away from it)
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A sled of mass m is given a kick on a frozen pond. The kick imparts to the sled an initial speed of 2.00 m/s . The coefficient o
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2.04 meters distance is traveled by the sled before stopping.

Mass of the sled = m

The initial speed of the sled = 2 m/s

Coefficient of kinetic friction between sled and ice = 0.100

Let the distance the sled moves before it stops be d.

Gravity = 9.8 m/ s²

Let the initial kinetic energy sled be

= K _{i}

K_{i} =  \frac{1}{2} mv ^{2}

The work done by the frictional force is,

Work \: done \:  by \: frictional \: force =W_{f}

W _{f} = μ_{k}mgd

Work done by frictional force= Initial kinetic energy of the sled

W_{f} = K_{i}

μ_{k}mgd= \frac{1}{2} mv ^{2}

So, the distance traveled by the sled before stopping is

d= \frac{1mv ^{2} }{2 \:μ_{k}mg}

d= \frac{1v ^{2} }{2 \:μ_{k}g}

d= \frac{2^{2} }{2  \times \:0.100 \times 9.8}

d= 2.04 \: m

Therefore, the distance traveled by the sled before stopping is 2.04 meters.

To know more about work done, refer to the below link:

brainly.com/question/13662169

#SPJ4

5 0
2 years ago
A cart traveling at 0.3 m/s collides with stationary object. After the collision, the cart rebounds in the opposite direction. T
miv72 [106K]

Answer:

C.   In the first collision has twice the momentum as when it stays still ( second colllions)

Explanation:

To see which statement is correct, it is best to solve the problem, the momentum is equal to the variation of the moment

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Case 2. The expensive one that still after the crash so its speed is zero (vf = 0)

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Let's calculate the relationship between the two impulses

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When it bounces it has twice the momentum as when it stays still

Now let's analyze the answers:

A.   False The momentum changes

B. False. The momentum is less in the second collision

C. True.  The momentum is double in this collision

D. False. Can be calculated, because the mass is the same throughout the exercise and is eliminated in the equations

E. False.  When they say bounces it implies the same speed with the opposite direction

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3 years ago
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