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swat32
2 years ago
11

All of the following are reasons why astronomers have difficulty looking at distant stars except: A. Some stars are so far away,

even the best telescopes are not powerful enough. B. They don't understand how redshift affects stars. C. There is a lot of gas and dust in the atmosphere that distorts the light coming from stars. D. There is a lot of light pollution on Earth.
Physics
1 answer:
dangina [55]2 years ago
3 0

Answer: Option B: They don't understand how redshifts affects stars.

There are millions and billions of stars in our universe. Scientists study stars using telescopes. But some stars are so far, that even the best powerful telescopes are not able to detect their light. Also, the interstellar distance is filled with clouds of dust and gas which makes some part of the universe opaque. It becomes difficult to study stars using earth based telescopes due to lot of light pollution. Astronomers understand how redshifts affects stars and hence, are able use this to study stars via their spectrum. Hence, astronomers don't understand how redshifts affects stars is a wrong reason of why they face difficulty in looking at distant stars.

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A trolley is moving at constant speed in a friction compensated track. Some plasticine is dropped on the trolley and sticks on i
Maurinko [17]

Answer:

Since the momentum of the body remains constant ( conserved) the trolley slows down (its velocity reduces) since its mass increases.

5 0
1 year ago
What if m1 is initially moving at 3.4 m/s while m2 is initially at rest? (a) find the maximum spring compression in this case?
Lisa [10]
<span>Ans : Initial E = KE = ½mv² = ½ * 1.2kg * (2.2m/s)² = 2.9 J max spring compression where both velocities are the same: conserve momentum: 1.2kg * 2.2m/s = (1.2 + 3.2)kg * v → v = 0.6 m/s which means the combined KE = ½ * (1.2 + 3.2)kg * (0.6m/s)² = 0.79 J The remaining energy went into the spring: U = (2.9 - 0.79) J = 2.1 J = ½kx² = ½ * 554N/m * x² x = 0.0076 m ↠(a)</span>
7 0
3 years ago
When the displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position, what fraction of the
KonstantinChe [14]

Answer:

The ratio is  KE : TM  =  0.75

Explanation:

from the question we are told that

  The displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position

Generally the total mechanical  energy of the mass is mathematically represented as

        TM  =  \frac{1}{2}  *  k  *  A^2

Here  k is the spring constant  ,  A is the total displacement of the  the mass  from maximum  compression to maximum extension of the spring

Generally this total mechanical energy is mathematically represented as

        TM  =  KE  + PE

=>     KE = TM  - PE

Here the potential  energy of the mass is mathematically represented as

     PE   = \frac{1}{ 2}  *  k *  [ x ]^2

Here x is the displacement of the mass from maximum compression or extension of the spring to equilibrium position and the value is  

      x = \frac{A}{2}

So

     PE   = \frac{1}{ 2}  *  k *  [ \frac{A}{2}  ]^2

So

      KE =  \frac{1}{2}  *  k  *  A^2 - \frac{1}{2}  *  k  *  [\frac{A}{2} ]^2

=>    KE =  \frac{1}{2}  *  k  *  A^2 - \frac{1}{8}  *  k  *  A ^2

=>    KE =  0.375  *  k  *  A^2

So the ratio of  KE :  TM is  mathematically represented as

       \frac{KE}{TM} =  \frac{0.375  k A^2 }{0.5 k A^2}

=>    \frac{KE}{TM} = 0.75

3 0
3 years ago
You are standing on a large sheet of frictionless ice and holding a large rock. In order to get off the ice, you throw the rock
kondor19780726 [428]

Answer:

0.4778 m/s

Explanation:

To solve this question, we will make use of law of conservation of momentum.

We are given that the rock's velocity is 12 m/s at 35°. Thus, the horizontal component of this velocity is;

V_x = (12 m/s)(cos(35°)) = 9.83 m/s.

Thus, the horizontal component of the rock's momentum is;

(3.5 kg)(9.83 m/s) = 34.405 kg·m/s.

Since the person is not pushed up off the ice or down into it, his momentum will have no vertical component and so his momentum will have the same magnitude as the horizontal component of the rock's momentum.

Thus, to get the person's speed, we know that; momentum = mass x velocity

Mass of person = 72 kg and we have momentum as 34.405 kg·m/s

Thus;

34.405 = 72 x velocity

Velocity = 34.405/72

Velocity = 0.4778 m/s

6 0
3 years ago
A ship leaves the island of Guam and sails a distance 255 km at an angle 49.0 o north of west. Part A: In which direction must i
kiruha [24]

Answer:

Explanation:

We shall represent displacement in vector form .Consider east as x axes and north as Y axes west as - ve x axes and south as - ve Y axes . 255 km can be represented by the following vector

D₁ = - 255 cos 49 i  + 255 sin49 j

= - 167.29 i + 192.45 j

Let D₂ be the further displacement which lands him 125 km east . So the resultant displacement is

D = 125 i

So

D₁ + D₂ = D

- 167.29 i + 192.45 j + D₂ = 125 i

D₂ = 125 i + 167.29 i - 192.45 j

= 292.29 i - 192.45 j

Angle of D₂ with x axes θ

tan θ = -192.45 / 292.29

= - 0.658

θ = 33.33 south of east

Magnitude of D₂

D₂² = ( 192.45)² + ( 292.29)²

D₂ = 350 km approx

Tan

7 0
3 years ago
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